$$$$

31  Theory

Highlights of this Chapter: we prove that the Riemann/Darboux integral satisfies our axioms of integration, and thus is truly an integral. We then use this construction to prove useful facts about the integral: includeing that the integral is a linear map, and power series can be integrated term by term within their radius of convergence.

Remark 31.1. Because the definitions of the Riemann and Darboux integral are equivalent, we are free to use whichever we please in deriving properties of this integral. While each definition has its own strengths, the formulation of Darboux will provide shorter proofs the majority of the time.

31.1 Verifying the Axioms

Theorem 31.1 (The Darboux Integral and Constants) Let $f(x)=k$ be a constant function, and $[a,b]$ an interval. Then $k$ is Darboux integrable on $[a,b]$ and $${\int }_{[a,b]}kdx=k(b-a)$$

Proof. For any partition $P$, we have $$M_i=\underset{x\in P_i}{sup}\{f(x)\}=k=infx\in P_i\{f(x)\}=m_i$$ as $f$ is constant. Thus, $$U(f,P)=\sum _{P_i\in P}{M}_i|P_i|=k\sum _{P_i\in P}|P_i|=k(b-a)$$ $$L(f,P)=\sum _{P_i\in P}{m}_i|P_i|=k\sum _{P_i\in P}|P_i|=k(b-a)$$ The upper and lower sums are constant, independent of partition, and so their respective infima/suprema are also constant, equal to this same value. Thus $k$ is integrable, and the integral is also this common value $${\int }_{[a,b]}kdx=k(b-a)$$

Theorem 31.2 (The Darboux Integral and Inequality) Let $f,g$ be Darboux integrable functions on $[a,b]$ and assume that $f(x)\le g(x)$ for all $x\in [a,b]$. Then $${\int }_{[a,b]}fdx\le {\int }_{[a,b]}gdx$$

Proof. The constraint $f\le g$ implies that on any partition $P$ we have $$L(f,P)\le L(g,P)$$ Or, equivalently $L(g,P)-L(f,P)\ge 0$. Taking the supremum over all $P$ of this set of nonnegative numbers yields a nonnegative number, so

$$\underset{P\in {\mathcal{P}}_{[a,b]}}{sup}\{L(g,P)-L(f,P)\}\ge 0$$ $$L(g)-L(f)\ge 0⟹L(f)\le L(g)$$ But since we’ve assumed $f$ and $g$ are integrable we know that $L(f)=U(f)={\int }_{a,b}fdx$ and $L(g)=U(g)={\int }_{[a,b]}gdx$. Thus

$${\int }_{[a,b]}fdx\le {\int }_{[a,b]}gdx$$

Theorem 31.3 (The Darboux Integral and Subdivision) Let $[a,b]$ be an interval and $c\in (a,b)$. Then a function $f$ defined on $[a,b]$ is Darboux-integrable on this interval if and only if it is Darboux integrable on both $[a,c]$ and $[c,b]$. Furthermore, when defined these three integrals satisfy the identity $${\int }_{[a,b]}fdx={\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx$$

Proof. First, assume that $f$ is integrable on $[a,b]$. By Theorem 29.1, this means for any $ϵ>0$ there exists a partition $P$ where $U(f,P)-L(f,P)<ϵ$. Now consider the refinement $P_c=P\cup \{c\}$. By the refinement lemma, $$L(f,P)\le L(f,P_c)\le U(f,P_c)\le U(f,P)$$ Thus $U(f,P_c)-L(f,P_c)<ϵ$ as well. Next we take this partition and divide it into partitions of each subinterval $P_1=P_c\cup [a,c]$ and $P_2=P_c\cup [c,b]$. By simply re-grouping the finite sums, we see $$L(f,P_c)=L(f,P_1)+L(f,P_2)U(f,P_c)=U(f,P_1)+U(f,P_2)$$

And, by the definitions of upper and lower sums, for each we know $U(f,P_i)-L(f,P_i)\ge 0$. All that remains to insure the integrability of $f$ on $[a,c]$ and $[c,b]$ is to show that these differences are individually less than $ϵ$. But this is immediate, as for example,

$$\begin{array}{rl}U(f,P_1)-L(f,P_1)& \le U(f,P_1)-L(f,P_1)+(U(f,P_2)-L(f,P_2))\\ & =(U(f,P_1)+U(f,P_2))-(L(f,P_1)+L(f,P_2))\\ & =U(f,P_c)-L(f,P_c)\\ & \le ϵ\end{array}$$

and the same argument applies to $U(f,P_2)-L(f,P_2)$.

Next we assume integrability on the two subintervals, and prove integrability on the whole interval.

Proof. Let $ϵ>0$ and by our integrability assumptions choose partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$ such that $$U(f,P_i)-L(f,P_i)\le \frac{ϵ}{2}i\in \{1,2\}$$ Now, their union $P=P_1\cup P_2$ is a partition of $[a,b]$, and re-grouping the finite sums, we see $$L(f,P)=L(f,P_1)+L(f,P_2)U(f,P)=U(f,P_1)+U(f,P_2)$$

Thus,

$$\begin{array}{rl}U(f,P)-L(f,P)& =(U(f,P_1)+U(f,P_2))-(L(f,P_1)+L(f,P_2))\\ & =(U(f,P_1)-L(f,P_1))+(U(f,P_2)-L(f,P_2))\\ & \le \frac{ϵ}{2}+\frac{ϵ}{2}\\ & =ϵ\end{array}$$

So, we see that integrability on $[a,b]$ is equivalent to integrability on $[a,c]$ and $[c,b]$. Finally, we need to show in the case where all three integrals are defined, the subdivision equality actually holds.

Proof. Let $P$ be any partition of the interval $[a,b]$ and define the usual suspects: $$P_c=P\cup \{c\}{P}_1=P_c\cup [a,c]P_2=P_c\cup [c,b]$$ We need three pieces of data. First, the inequalities relating integrals to upper and lower sums $$L(f,P_1)\le {\int }_{[a,c]}fdx\le U(f,P_1)L(f,P_2)\le {\int }_{[c,b]}fdx\le U(f,P_2)$$ Second, the inequalities of refinements: $$L(f,P)\le L(f,P_c)\le U(f,P_c)\le U(f,P)$$ and third, the relationships between $P_1,P_2$ and $P_c$: $$L(f,P_c)=L(f,P_1)+L(f,P_2)U(f,P_c)=U(f,P_1)+U(f,P_2)$$

Putting all of these together, we get both lower and upper estimates for the sum of the integrals over the subdivision:

$$L(f,P)\le L(f,P_c)=L(f,P_1)+L(f,P_2)\le {\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx$$ $${\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx\le U(f,P_1)+U(f,P_2)=U(f,P_c)\le U(f,P)$$

And concatenating these inequalities gives the overall bound, for any arbitrary partition $P$:

$$L(f,P)\le {\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx\le U(f,P)$$

Thus, the sum of these integrals lies between the upper and lower sum of $f$ on $[a,b]$ for every partition. As $f$ is integrable, we know there is a single number with this property, and that number is by definition the integral. Thus $${\int }_{[a,b]}fdx={\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx$$

Phew! We’ve successfully verified all three axioms for the Darboux integral. Taken together, these prove that our construction really is an integral!

Corollary 31.1 The equality of upper and lower sums satisfies the axioms of integration, and thus the Darboux Integral really does define an integral.

31.2 Integrability

Now, we show that our constructed integral is actually interesting - that all continuous functions are integrable!

Theorem 31.4 (Continuous $⟹$ Integrable) Every continuous function on a closed interval is Darboux integrable.

Proof. Let $f$ be continuous on the interval $[a,b]$ and choose $ϵ>0$. We will prove integrability by finding a partition $P$ such that $U(f,P)-L(f,P)<ϵ$.

As $f$ is continuous it is bounded (by the extreme value theorem), so the upper and lower sums are defined for all partitions. It is also uniformly continuous (as $[a,b]$ is a closed interval), so we can find a $\delta$ such that $$|x-y|<\delta ⟹|f(x)-f(y)|<\frac{ϵ}{b-a}$$

Now, choose a partition $P$ of $[a,b]$ where the width of each interval is less than $\delta$. Comparing upper and lower sums on this interval, $$U(f,P)-L(f,P)=\sum _{P_i\in P}{M}_i|P_i|-\sum _{P_i\in P}{m}_i|P_i|=\sum _{P_i\in P}[M_i-m_i]|P_i|$$

Since $|P_i|<\delta$, we know that for any $x,y\in P_i$ the values $f(x),f(y)$ differ by less than $ϵ/(b-a)$. Thus the difference of between the infimum and supremum over this interval must be less than or equal to this bound:

$$M_i-m_i\le \frac{ϵ}{b-a}$$

Using this to bound our sum, we see

$$U(f,P)-L(f,P)=\sum _{P_i\in P}[M_i-m_i]|P_i|\le \frac{ϵ}{b-a}\sum _{P_i\in P}|P_i|$$ $$=\frac{ϵ}{b-a}(b-a)=ϵ$$

Thus, $f$ is integrable!

But the Darboux integral allows us to integrate even more things than the continuous functions. For example, it is quite straightforward to prove that all monotone functions are integrable (even those with many discontinuities!)

Theorem 31.5 (Monotone $⟹$ Integrable) Every monotone bounded function on a closed interval is integrable.

Proof. Without loss of generality let $f$ be monotone increasing and bounded on the interval $[a,b]$ and choose $ϵ>0$. We will prove integrability by finding a partition $P$ such that $U(f,P)-L(f,P)<ϵ$.

Let $B=f(b)-f(a)$ be the difference between values of $f$ at the endpoints. If $B=0$ then $f$ is constant, and we already know constant functions are integrable so we are done.

Otherwise, let $P$ be an arbitrary evenly spaced partition of widths $\mathrm{\Delta }=ϵ/B$, we consider the difference $U(f,P)-L(f,P)$:

$$U(f,P)-L(f,P)=\sum _{P_i\in P}{M}_i|P_i|-\sum _{P_i\in P}{m}_i|P_i|$$ $$=\sum _{P_i\in P}[M_i-m_i]|P_i|=\mathrm{\Delta }\sum _{P_i\in P}[M_i-m_i]$$

Since $f$ is increasing, its supremum on each interval occurs on the right, and its infimum on the left. That is, if $P_i=[t_{i-1},t_i]$ we have $$m_i=f(t_{i-1})M_i=f(t_i)$$ Plugging this into the above gives a telescoping sum! $$U(f,P)-L(f,P)=\mathrm{\Delta }\sum _{1\le i\le n}[f(t_i)-f(t_{i-1})]=\mathrm{\Delta }[f(t_n)-f(t_0)]$$ But $t_0=a$ and $t_n=b$ are the endpoints of our partition, and so this equals

$$=\mathrm{\Delta }[f(b)-f(a)]=\frac{ϵ}{f(b)-f(a)}[f(b)-f(a)]=ϵ$$

And, inductively its straightforward to show (via subdivision) that if the domain of a function can be partitioned into finitely many intervals on which it is integrable, than its integrable on the entire thing. Thus, for example piecewise continuous functions are Darboux Integrable. The precise statement and theorem is below.s

Definition 31.1 (Piecewise Integrable Function) A function $f$ defined on a domain $I$ is piecewise integrable if $I$ is the disjoint of a finite sequence of intervals $I=I_1\cup I_2\cup \dots \cup I_n$, and $f$ restricted to each interval is integrable.

Proposition 31.1 (Piecewise Integrable $⟹$ Integrable) If $f$ is piecewise integrable, then it is integrable.

Proof. We begin with the case that $f$ is piecewise integrable on two subintervals, $[a,c]$ and $[c,b]$ of the interval $[a,b]$. Then the subdivision axiom immediately implies that $f$ is in fact integrable on the entire interval.

Now, assume for induction that all functions that are piecewise integrable on intervals with $\le n$ subdivisions are actually integrable, and let $f$ be a piecewise integrable function on a union of $n+1$ intervals $$[a,b]=I_1\cup I_2\cup \cdots \cup I_n\cup I_{n+1}$$

Set $J$ equal to the union of the first $n$, so that $[a,b]=J\cup I_{n+1}$. Then when restricted to $J$, the function $f$ is piecewise integrable on $n$ intervals, so its integrable by assumption. And so, $f$ is integrable on both $J$ and $I_{n+1}$, so its piecewise integrable with two intervals, and hence integrable as claimed.

Because all continuous functions and all monotone functions are integrable, we have the following useful corollary covering most functions usually seen in a calculus course.

Corollary 31.2 All piecewise continuous and piecewise monotone functions with finitely many pieces are integrable.

But monotone functions are even more general than this! A monotone function can have countably many discontinuities. Pursuing this further, there is a precise characterization of Darboux-Integrable functions (which we state, but do not prove).

Theorem 31.6 (Riemann-Lebesgue Integrability Theorem) A function $f$ on the interval $[a,b]$ is Riemann/Darboux-integrable if and only if it is bounded and its set of discontinuities is measure zero.

We do not prove this theorem here (its proof is long, and requires a precise definition of the concept of a measure zero set to even state) nor do we use it in what follows. But it is interesting to note that if one were to prove this theorem, many of the results both above and below follow as rather trivial consequences:

• Continuous functions are bounded (the extreme value theorem) and have empty discontinuity set. Thus they are integrable.
• One can prove that monotone functions can have at most countably many discontinuities, and any countable set has measure zero. Thus monotone functions are integrable.
• Piecewise integrability implies integrability as the overall function is bounded by the max of the bounds on each interval, and the overall discontinuity set is the union of the discontinuity sets (and, the finite union of measure zero sets has measure zero).
• Constant multiples of integrable functions are integrable: multiplying a bounded function by a constant leaves it still bounded, and does not change the discontinuity set.
• Sums of integrable functions are integrable: a sum is bounded by the sum of the bounds for its terms, and its discontinuity set is contained in the union of the discontinuity sets of each term.

31.3 Linearity

Theorem 31.7 (Integrability of Multiples) Let $f$ be an integrable function a closed interval $I$, and $c\in \mathbb{R}$. Then the function $cf$ is also integrable on $I$, and furthermore $${\int }_{I}cfdx=c{\int }_{I}fdx$$

Proof. Let $P_n$ be an arbitrary sequence of shrinking partitions (of length $N_n$), and $S_n$ an arbitrary sequence of sample points for $P_n$. We attempt to evaluate the limit

$$lim\sum _I(cf,P_n,S_n)$$

For any partition $P$ and sample set $S$, $\sum _I(cf,P,S)$ is a finite sum, and so we may factor out the constant $c$:

$$\begin{array}{rl}\sum _I(cf,P,S)& =\sum _{0\le k<N}cf(s_k)|P_k|\\ & =c\sum _{0\le k<N}f(s_k)|P_k|\\ & =c\sum _I(f,P,S)\end{array}$$

Doing this for each $n$ yields

$$lim\sum _I(cf,P_n,S_n)=limc\sum _I(f,P_n,S_n)$$

Since $f$ is assumed integrable on $I$ we know that $lim\sum _I(f,P_n,S_n)={\int }_{I}fdx$ and so we may use the limit theorems to pull the constant $c$ out:

$$limc\sum _I(f,P_n,S_n)=clim\sum _I(f,P_n,S_n)=c{\int }_{I}fdx$$

Thus, $lim\sum _I(cf,P_n,S_n)$ exists, and as $P_n,S_n$ were arbitrary, its value is independent of the particular choice of shrinking partitions. By definition this means that $cf$ is integrable on $I$, and that

$${\int }_{I}cfdx=c{\int }_{I}fdx$$

Exercise 31.1 For practice, provide a proof of this using the Darboux integrability criterion, instead of the definition of the Riemann integral.

We proceed with the same strategy to prove the integrability of a sum of integrable functions: using the limit laws and definitions for finite sums to calculate along the way:

Theorem 31.8 (Integrability of Sums) Let $f,g$ be integrable functions on a closed interval $I$. Then their sum $f+g$ is also integrable on $I$. Furthermore, its integral is the sum of the integrals of $f$ and $g$: $${\int }_I(f+g)={\int }_{I}f+{\int }_{I}g$$

Proof. Let $P_n$ be an arbitrary sequence of shrinking partitions, and for each $n$, let $S_n$ be a sample set for $P_n$. We attempt to evaluate the limit $$lim\sum _I(f+g,P_n,S_n)$$ For any partition $P$ and sample $S$, $\sum _I(f+g,P,S)$ is a finite sum, and so we can re-order its terms by the commutativity of addition, decomposing into two Riemann sums

$$\begin{array}{rl}\sum _I(f+g,P,S)& =\sum _{0\le k<N}[f(s_k)+g(s_k)]|P_k|\\ & =\sum _{0\le k<N}f(s_k)|P_k|+\sum _{0\le k<N}g(s_k)|P_k|\\ & =\sum _I(f,P,S)+\sum _I(g,P,S)\end{array}$$

Doing this for each $n$ yields

$$lim\sum _I(f+g,P_n,S_n)=lim[\sum _I(f,P_n,S_n)+\sum _I(g,P_n,S_n)]$$

By hypothesis, both $f$ and $g$ are integrable on $I$, meaning that $$lim\sum _I(f,P_n,S_n)={\int }_{I}fdxlim\sum _I(g,P_n,S_n)={\int }_{I}gdx$$

Since both of these limits exist, we can use the limit law for sums to distribute the limit above, and see

$$\begin{array}{rl}lim\sum _I(f+g,P_n,S_n)& =lim[\sum _I(f,P_n,S_n)+\sum _I(g,P_n,S_n)]\\ & =lim\sum _I(f,P_n,S_n)+lim\sum _I(g,P_n,S_n)\\ & ={\int }_{I}fdx+{\int }_{I}gdx\end{array}$$

Since $P_n$ and $S_n$ were arbitrary, this same result must hold for all such shrinking partitions - all such limits converge, and have the same value ${\int }_{I}fdx+{\int }_{I}gdx$. Thus, by definition $f+g$ is integrable, and

$${\int }_{I}f+gdx={\int }_{I}fdx+{\int }_{I}gdx$$

Exercise 31.2 For practice, provide a proof of this using the Darboux integrability criterion, instead of the definition of the Riemann integral.

Each of these theorems does two things: it proves something about the space of integrable functions and also about how the integral behaves on this space. Below we rephrase the conclusion of these theorems in the terminology of linear algebra - a result so important it deserves the moniker of “Theorem” itself.

Theorem 31.9 (Linearity of the Riemann/Darboux Integral) For each interval $[a,b]\subset \mathbb{R}$, the set $\mathcal{I}([a,b])$ of Riemann integrable functions forms a Vector Subspace of the set of all functions $[a,b]\to \mathbb{R}$. On this subspace, the Riemann integral defines a linear map

$${\int }_{[a,b]}:\mathcal{I}([a,b])\to \mathbb{R}$$

31.4 Power Series

We now turn to the issue of integrating power series. The theoretical results are in close analogy to the differentiation case: in summary,

• A power series is integrable on the interior of its radius of convergence, and the the antiderivative converges on the same interval
• The antiderivative can be found by antidifferentiating term-by-term.

There are two ways we could try to prove a theorem such as this: one, we could try to mimic the style of the differentiation proof, developing a theory of dominated convergence for the Riemann integral. This succeeds without issue, and is carried out in the following section. But alternatively we could attempt to use the fundamental theorem of calculus to relate this directly to what we already know about differentiation. This turns out to be a shorter and more elementary argument, and while less general (it applies only to power series, not to general series of functions) it is more than sufficient for our purposes, so we give it here.

Theorem 31.10 Let $f(x)=\sum _{n\ge 0}{a}_n{x}^n$ be a power series with radius of convergence $R$. Then the power series $F(x)=\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}$ has the same radius of convergence, and $$F(x)={\int }_{[a,x]}fdx$$

We will prove this theorem as a sequence of propositions. First, we check that the radius of convergence remains unchanged under term-by-term integration of a series.

Proof. Like for the differentiable case, we prove this here under the assumption that the Ratio test succeeds in computing the radius of convergence for the original series, so for any $x\in (-R,R)$ $$lim\left|\frac{a_{n+1}}{a_n}\right||x|<1$$

We now turn to compute the ratio test for our new series $\sum _{\frac{a_n}{n+1}}{x}^{n+1}$: the ratio in question is

$$\frac{\frac{a_{n+1}}{n+2}{x}^{n+2}}{\frac{a_n}{n+1}{x}^{n+1}}=\left(\frac{a_{n+1}}{a_n}\right)\left(\frac{n+1}{n+2}\right)x$$

Since $(n+1)/(n+2)\to 1$ we can compute the overall limit using the limit theorems and see we end up with the exact same limit as for the original series! Thus integrating term by term does not change the radius of convergence at all.

Now, we turn to the main event: we prove that the term-by-term antiderivative is the integral of the original power series.

Proof. Let $f(x)=\sum _{n\ge 0}{a}_n{x}^n$ have radius of convergence $R$, and let $F(x)=\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}$. Choose any $x\in (0,R)$; we wish to show that $$F(x)={\int }_{[0,x]}fdx$$

Because term-by-term-differentiation holds within the radius of convergence, for any $|x|<R$ we have $$\begin{array}{rl}{F}^{\prime }(x)& ={\left(\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}\right)}^{\prime }\\ & =\sum _{n\ge 0}{\left(\frac{a_n}{n+1}{x}^{n+1}\right)}^{\prime }\\ & =\sum _{n\ge 0}\frac{a_n}{n+1}(n+1)x^n\\ & =\sum _{n\ge 0}{a}_n{x}^n\\ & =f(x)\end{array}$$

Thus, $F$ is an antiderivative of $f$! So by the fundamental theorem of calculus, we know that we can use $F$ to evaluate the integral:

$${\int }_{[0,x]}fdx=F(x)-F(0)$$

But $F(0)=0$ as $F$ has no constant term! Thus we have it,

$${\int }_{[0,x]}\sum _{n\ge 0}{a}_n{x}^{n}dx=\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}$$

31.5 $★$ Dominated Convergence

Now we turn to the development of a more general theory, based on a new Dominated Convergence Theorem for the Riemann integral.

Theorem 31.11 (Dominated Convergence for the Riemann/Darboux Integral) Let $\{f_n\}$ be a sequence of Riemann integrable functions on a closed interval $I$, and assume that the functions $f_n$ converge pointwise to a Riemann integrable function $f$. Then if there exists some $M$ where $|f_n(x)|<M$ for all $x\in I$, the order of integration and limit may be interchanged: $$lim{\int }_I{f}_n={\int }_{I}f$$

Proof. For a short elementary proof, see https://arxiv.org/pdf/1408.1439.pdf.

Looking back at the original Dominated Convergence for sums, there are some slight differences here: we have to assume the integrability of the limit, and we must give a uniform bound on all the $f_n$. However, if we are mindful of these differences the application of the theorem is analogous - we use it to switch a sum and integral, and then integrate term by term.

Remark 31.2. One motivating reason to seek an alternative theory of integration in advanced analysis is to find an integral with a dominated convergence theorem closer to the others we’ve met. Such an integral exists, and was first constructed by Henri Lebesgue in 1905 (but will not concern us here; dominated convergence for the Riemann integral is plenty powerful!)

Again, both series and integrals are defined by limit statements, so this proof is going to require an interchange of limits. Instead of digging all the way down to the foundations and using the Riemann sum definitions to apply Dominated convergence for series (Theorem 21.1), we can save some trouble by directly using Dominated convergence for integrals (Theorem 31.11)

Just as before we first show how the proof goes assuming that we can exchange the sum and integral limits, and then we’ll justify that this is allowed:

Theorem 31.12 (Integration of Power Series) Let $f=\sum _{k\ge 0}{a}_k{x}^k$ be a power series with radius of convergence $R$. Then for $x\in (-R,R)$: $${\int }_{[0,x]}f=\sum _{k\ge 0}\frac{a_k}{k+1}{x}^{k+1}$$

Proof. Let $f_N$ denote the $N^{th}$ partial sum of the series, $f_N=\sum _{k=0}^N{a}_k{x}^k$, so $f=\underset{N}{lim}{f}_N$. Substituting this into the above, $${\int }_{[0,x]}f={\int }_{[0,x]}\underset{N}{lim}{f}_N$$ Now assuming that dominated convergence for integrals applies, we may switch the integral and limit statement, to get $${\int }_{[0,x]}\underset{N}{lim}{f}_N=\underset{N}{lim}{\int }_{[0,x]}{f}_N$$ Now, each $f_N$ is a polynomial - meaning its a finite sum! This means we can integrate it term by term using the linearity of the integral (Theorem 31.8):

$$\begin{array}{rl}{\int }_{[0,x]}{f}_N& ={\int }_{[0,x]}\sum _{k=0}^N{a}_k{t}^k\\ & =\sum _{k=0}^N{a}_k{\int }_{[0,x]}{t}^k\\ & =\sum _{k=0}^N{a}_k\frac{x^{k+1}}{k+1}\end{array}$$

Now, taking the limit $N\to \mathrm{\infty }$ gives the series of term by term antiderivatives:

$${\int }_{[0,x]}f=\underset{N}{lim}\sum _{k=0}^N{a}_k\frac{x^{k+1}}{k+1}=\sum _{k\ge 0}\frac{a_k}{k+1}{x}^{k+1}$$

Now, we need to justify that dominated convergence applies. Theorem 31.11 requires two things: (1) that the limit $lim{f}_N=f$ is integrable on $[0,x]$, and (2) that each of the functions $f_N$ is uniformly bounded by some constant $M$ on the interval $[0,x]$.

Proposition 31.2 If $f$ is a power series and $x$ is within the radius of convergence, then $f$ is integrable on $[0,x]$.

Proof. If $x\in (-R,R)$ then the closed interval $[0,x]$ is completely contained within the interval of convergence. Because a power series is continuous at each point on the interior of its interval of convergence (Theorem 21.4), it is continuous on the closed interval $[0,x]$.
And, as continuous functions on a closed interval are integrable (Theorem 31.4), it is integrable on $[0,x]$ as required.

The second requirement requires us to dig into the definition of a power series a bit.

Proposition 31.3 Let $f$ be a power series with radius of convergence $R$, and $f_N$ be its sequence of partial sums. Then if $x\in (0,R)$, there is a fixed constant $M$ such that $$|f_N(t)|\le M\mathrm{\forall }N\mathrm{\forall }t\in [0,x]$$

Proof. As $0<x<R$ the interval $[0,x]$ is contained in the interior of the interval of convergence, so the power series $f=\sum _{k\ge 0}{a}_k{t}^k$ is absolutely convergent for each $t\in [0,x]$. Let $g$ denote the series of term-wise absolute values $g(t)=\sum _{k\ge 0}|a_k|t^k$, and $g_N$ denote its sequence of partial sums. Then, by the triangle inequality for finite sums, for every $t\in [0,x]$, $$|f_N(t)|=\left|\sum _{k=0}^N{a}_k{t}^k\right|\le \sum _{k=0}^N|a_k|t^k=g_N(t)$$ And, since all the terms of $g$ are positive, the sequence $g_N(t)$ is monotone increasing in $N$, with $$g_N(t)\le g(t)\mathrm{\forall }N$$ Stringing these two inequalities together, we see that for each $t\in [0,x]$, the quantity $g(t)$ is an upper bound for $\{f_N(t)\}$.

But $g$ itself is a power series (with coefficients $|a_k|$) and is convergent for all $t\in [0,x]$ (as $f$ is absolutely convergent at all points on the interior of its radius of convergence). Thus by Theorem 21.4, $g$ is continuous on $[0,x]$. That means we can apply the extreme value theorem (Theorem 16.1) to find an absolute maximum of $g$ on $[0,x]$: a value $M$ such that $g(t)\le M$ for all $t\in [0,x]$.

Now truly stringing it all together, we see that for each $t\in [0,x]$ and each $N\in \mathbb{N}$, $$|f_N(t)|\le g_N(t)\le g(t)\le M$$ Thus $M$ is the uniform bound we seek.

Remark 31.3. Note we could get by here without invoking the extreme value theorem,but rather just to see $\{g(t)\mid t\in [0,x]\}$ is a bounded set, and select $M$ to be any upper bound. We chose the (stronger) extreme value theorem only because it is more memorable.

Exercise 31.3 Use the argument above to show that this holds for any $x\in (-R,R)$; the assumption on positivity is not required.