31 Integrable Functions
31.1 Integrability
Theorem 31.1 (Continuous functions are Integrable) Every continuous function on a closed interval is Darboux integrable.
Proof. Let \(f\) be continuous on the interval \([a,b]\) and choose \(\epsilon>0\). We will prove integrability by finding a partition \(P\) such that \(U(f,P)-L(f,P)<\epsilon\).
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 \[|x-y|<\delta\implies |f(x)-f(y)|<\frac{\epsilon}{b-a}\]
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 \(\epsilon/(b-a)\). Thus the difference of between the infimum and supremum over this interval must be less than or equal to this bound:
\[M_i-m_i\leq \frac{\epsilon}{b-a}\]
Using this to bound our sum, we see
\[U(f,P)-L(f,P)=\sum_{P_i\in P}[M_i-m_i]|P_i|\leq \frac{\epsilon}{b-a}\sum_{P_i\in P}|P_i|\] \[=\frac{\epsilon}{b-a}(b-a)=\epsilon\]
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 31.2 (Monotone functions are 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 \(\epsilon>0\). We will prove integrability by finding a partition \(P\) such that \(U(f,P)-L(f,P)<\epsilon\).
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 \(\Delta = \epsilon/B\), we consider the difference \(U(f,P)-L(f,P)\):
\[U(f,P)-L(f,P)=\sum_{P_i\in P}M_i|P_i|-\sum_{P_i\in P}m_i|P_i|\] \[=\sum_{P_i\in P}[M_i-m_i]|P_i|=\Delta \sum_{P_i\in P}[M_i-m_i]\]
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 \[m_i=f(t_{i-1})\hspace{1cm}M_i=f(t_i)\] Plugging this into the above gives a telescoping sum! \[U(f,P)-L(f,P)=\Delta\sum_{1\leq i \leq n}[f(t_i)-f(t_{i-1})]=\Delta [f(t_n)-f(t_0)]\] But \(t_0=a\) and \(t_n=b\) are the endpoints of our partition, and so this equals
\[=\Delta[f(b)-f(a)]=\frac{\epsilon}{f(b)-f(a)}[f(b)-f(a)]=\epsilon\]
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 Functions) A function \(f\) defined on a domain \(I\) is piecewise integrable if \(I\) is the disjoint of a finite sequence of intervals \(I=I_1\cup I_2\cup\ldots \cup I_n\), and \(f\) restricted to each interval is integrable.
Proposition 31.1 (Piecewise Integrable \(\implies\) 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 \(\leq n\) 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 31.1 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 31.3 (\(\bigstar\) Riemann’s Integrability Criterion) 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.
31.2 \(\blacklozenge\) Linearity
Theorem 31.4 (Integrability of Constant Multiples) Let \(f\) be an integrable function a closed interval \(I\), and \(c\in\RR\). Then the function \(cf\) is also integrable on \(I\), and furthermore \[\int_I cf=c\int_I f\]
We separate into cases depending on the sign of \(c\). Below we complete \(c\geq 0\), and leave \(c<0\) as an exercise.
Proof (\(c=0\)). When \(c=0\) the function \(cf\) is identically the zero function. Thus by Proposition 29.4 \[\int_I cf = \int_I 0 = 0|I|=0\] This is equal to \(c\int_I f = 0\int_I f = 0\), so we’ve proven \(c\int_I f = \int_I cf\) as required.
Proof (\(c>0\)). For \(c>0\), note that on any interval \(J\) we have \[\inf_{x\in J}\{cf(x)\}=c\inf_{x\in J}\{f(x)\}\hspace{1cm}\sup_{x\in J}\{cf(x)\}=c\sup_{x\in J}\{f(x)\}\] Thus for any partition \(P\), \[L(cf,P)=\sum_i \inf_{x\in P_i}\{cf(x)\}|P_i|=c\sum_i \inf_{x\in P_i}{f(x)}|P_i|=cL(f,P)\] \[U(cf,P)=\sum_i \sup_{x\in P_i}\{cf(x)\}|P_i|=c\sum_i \sup_{x\in P_i}{f(x)}|P_i|=cU(f,P)\]
Let \(P_n\) be any sequence of shrinking partitions: since \(f\) is integrable we know \(\lim U(f,P_n)=\lim L(f,P_n)=\int_I f\). Computing with the limit laws
\[ \lim L(cf, P_n)=\lim c L(f,P_n)=c\lim L(f,P_n)=c\int_I f\] \[ \lim U(cf, P_n)=\lim c U(f,P_n)=c\lim U(f,P_n)=c\int_I f\]
Thus the upper and lower sums are equal in the limit, so \(cf\) is integrable and its integral is equal to their common value \(c\int_I f\).
Exercise 31.1 Prove the \(c=-1\) case: if \(f\) is integrable on \(I\) then so is \(-f\) and \(\int_I(-f)=-\int_I f\). Hint: what does multiplying by \(-1\) do to \(m=\inf\) and \(M=\sup\) on each partition? What does it do to the sums \(U(f,P)\) and \(L(f,P)\)?
Combine this with the \(c>0\) case above to prove the analogous result for any negative constant multiple.
Theorem 31.5 (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. The key inequality bounding sums of functions on an arbitrary interval \(J\) is \[\inf_{x\in J}\{f(x)\}+\inf_{x\in J}\{g(x)\}\leq \inf_{x\in J} \{f(x)+g(x)\}\leq \sup_{x\in J} \{f(x)+g(x)\}\leq \sup_{x\in J}\{f(x)\}+\sup_{x\in J}\{g(x)\}\]
Given an arbitrary partition \(P\), summing over the subintervals \(P_i\) yields
\[L(f,P)+L(g,P)\leq L(f+g,P)\leq U(f+g,P)\leq U(f,P)+U(g,P)\]
By assumption \(f\) and \(g\) are both integrable, so we may select a sequence \(P_n\) of shrinking partitions such that \[\lim L(f,P_n)=\lim U(f,P_n)=\int_I f\hspace{1cm}\lim L(g,P_n)=\lim U(g,P_n)=\int_I g\]
Taking the limit of the above inequalities along this sequence of partitions yields
\[\int_I f + \int_I g \leq \lim L(f+g,P)\leq \lim U(f+g,P) \leq \int_I f+\int_I g\]
Thus by the squeeze theorem these limits are equal; so \(f+g\) is integrable, and its integral equals their common value \(\int_I f+\int_I g\).
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 31.6 (Linearity of the Riemann/Darboux Integral) For each interval \([a,b]\subset\RR\), the set \(\mathcal{I}([a,b])\) of Riemann integrable functions forms a Vector Subspace of the set of all functions \([a,b]\to\RR\). On this subspace, the Riemann integral defines a linear map
\[\int_{[a,b]}\colon\, \mathcal{I}([a,b])\to\RR\]
31.3 \(\blacklozenge\) Dominated Convergence for Integrals
Theorem 31.7 (Dominated Convergence for the 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: \[\lim \int_I f_n=\int_I f\]
Remark 31.1. This dominated convergence theorem is weaker than the form we are used to for sums. Previously, we have checked the dominating conditions and then concluded two things: (1) the resulting sum converges, and (2) we can interchange the sum and limit. For the Darboux integral, we need to demote (1) from a conclusion to an additional hypothesis.
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 Darboux integral is plenty powerful!)
Proof.
31.3.1 \(\oplus\) Integration of Power Series
This material will be re-done in a simpler way with FTC
PSERIES RADIUS OF CONVERGENCE: Forward cite to where the proof is.
Theorem 31.8 (Term-by-Term Integration of Power Series) Let \(f=\sum_{k\geq 0}a_kx^k\) be a power series with radius of convergence \(R\). Then for \(x\in(-R,R)\): \[\int_{[0,x]}f = \sum_{k\geq 0}\frac{a_k}{k+1}x^{k+1}\]
We give a proof using Dominated Convergence, followed by some lemmas justifying this applies to power series.
Proof. Let \(f_N\) denote the \(N^{th}\) partial sum of the series, \(f_N=\sum_{k=0}^N a_k x^k\), so \(f=\lim_N f_N\). Substituting this into the above, \[\int_{[0,x]}f=\int_{[0,x]}\lim_N f_N\] Now assuming that dominated convergence for integrals applies, we may switch the integral and limit statement, to get \[\int_{[0,x]}\lim_N f_N=\lim_N\int_{[0,x]}f_N\] 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 31.5):
\[\begin{align*} \int_{[0,x]}f_N&=\int_{[0,x]}\sum_{k=0}^N a_kt^k\\ &=\sum_{k=0}^N a_k\int_{[0,x]}t^k\\ &=\sum_{k=0}^N a_k \frac{x^{k+1}}{k+1} \end{align*}\]
Now, taking the limit \(N\to\infty\) gives the series of term by term antiderivatives:
\[\int_{[0,x]}f=\lim_N \sum_{k=0}^N a_k \frac{x^{k+1}}{k+1} =\sum_{k\geq 0}\frac{a_k}{k+1}x^{k+1}\]
Now, we need to justify that dominated convergence applies. Theorem 31.7 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 (?thm-power-series-continuity), it is continuous on the closed interval \([0,x]\).
And, as continuous functions on a closed interval are integrable (?thm-continuous-integrability), 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 \[|f_N(t)|\leq M\,\,\forall N\,\,\forall t\in[0,x]\]
Proof. As \(0<x<R\) the interval \([0,x]\) is contained in the interior of the interval of convergence, so the power series \(f=\sum_{k\geq 0}a_kt^k\) is absolutely convergent for each \(t\in[0,x]\). Let \(g\) denote the series of term-wise absolute values \(g(t)=\sum_{k\geq 0}|a_k|t^k\), and \(g_N\) denote its sequence of partial sums. Then, by the triangle inequality for finite sums, for every \(t\in[0,x]\), \[|f_N(t)|=\left|\sum_{k=0}^Na_kt^k\right|\leq \sum_{k=0}^N|a_k|t^k=g_N(t)\] And, since all the terms of \(g\) are positive, the sequence \(g_N(t)\) is monotone increasing in \(N\), with \[g_N(t)\leq g(t)\,\,\forall N\] 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 ?thm-power-series-continuity, \(g\) is continuous on \([0,x]\). That means we can apply the extreme value theorem (?thm-extreme-value) to find an absolute maximum of \(g\) on \([0,x]\): a value \(M\) such that \(g(t)\leq M\) for all \(t\in[0,x]\).
Now truly stringing it all together, we see that for each \(t\in[0,x]\) and each \(N\in\NN\), \[|f_N(t)|\leq g_N(t)\leq g(t)\leq M\] Thus \(M\) is the uniform bound we seek.
Exercise 31.2 Use the argument above to show that this holds for any \(x\in(-R,R)\); the assumption on positivity is not required.
31.4 \(\bigstar\) Order of Multiple Integrals
Exchanging limits! Conditions on when you can do this (constant bounds)
Maybe something about multiple bounds, and higher dimensional calculus?
Theorem 31.9 Let \(f(x,y)\) be a function such that
\[\int_I\int_J f = \int_J\int_I f\]
As a corollary we can offer a re-proof of the equality of mixed partials.
Corollary 31.2 Let \(f(x,y)\) be a function of two variables, such that the mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) exist and are continuous in both \(x\) and \(y\). Then they are equal.
Proof.
31.5 Differentiating Under the Integral
Theorem 31.10 Let \(f_t(x)\colon I\to \RR\) be a family of functions such that
- For each fixed \(t\), \(f_t\) is continuous on \(I\)
- For each fixed \(x\), the function \(t\mapsto f_t(x)\) is differentiable.
- The derivative \(\tfrac{d}{dt}f_t(x)\) is continuous on \(I\).
Then the function \(t\mapsto \int_I f_t\) is differentiable, and \[\frac{d}{dt} \int_I f_t = \int_I \frac{d}{dt}f_t\]
Proof. Expressing the derivative as a limit, we compute using the linearity of the integral
\[\begin{align*} \frac{d}{dt}\int_I f_t &= \lim_{h\to 0}\frac{\int_I f_{t+h} - \int_I f_t}{h}\\ &= \lim_{h\to 0} \frac{\int_I(f_{t+h}-f_t)}{h}\\ &=\lim_{h\to 0}\int_I\frac{f_{t+h}-f_t}{h} \end{align*}\]
This leaves us with a limit of a sequence of integrals, where we need to switch the order of the integral and the limit. To apply dominated convergence, select an arbitrary sequence \(h_n\to 0\) with \(h_n\neq 0\) and note
- For each \(n\), \((f_{t+h_n}-f_t)/h_n\) is continuous, and hence integrable.
- The function \(\tfrac{d}{dt}f_t\) is continuous, and hence integrable.
The final condition that needs to be checked is the existence of an \(M\) such that for all \(n\in\NN\) and all \(x\in I\), the difference quotient \((f_{t+h_n}(x)-f_t(x))/h_n\) is less than \(M\) in absolute value. For each \(n\), since \(|(f_{t+h_n}(x)-f_t(x))/h_n|\) is a continuous function of \(x\) on the closed interval \(I\), and so is bounded. If this collection is uniformly bounded by some fixed \(M\), we are done. So, assume for the sake of contradiction it is not, and for each \(k\in N\) there is some \(n_k\) and \(x_k\) where \[\left|\frac{f_{t+h_{n_k}}(x_k)-f_t(x_k)}{h_{n_k}}\right|>k\]
The points \(x_k\) lie in the closed interval \(I\), so by the Bolzano Weierstrass theorem we may select a convergent subsequence (to ease proliferation of indices, we will simply assume without loss of generality that \(x_k\) itself converges). Say \(x_k\to x^\star\in I\): then the limit above tending to infinity implies that \(\tfrac{d}{dt}f_t(x^\star)\) does not exist by the exercise below (it would be infinite): this is our contradiction, as we assumed \(\tfrac{d}{dt}f_t\) is continuous (and hence defined!).
Together with the previous points, this means dominated convergence applies! Switching the order completes the proof, as the resulting integrand is the definition of the \(t\) derivative of \(f_t\). \[\lim_{h\to 0}\int_I\frac{f_{t+h}-f_t}{h}=\int_I\lim_{h\to 0}\frac{f_{t+h}-f_t}{h}=\int_I \frac{d}{dt}f_t\]
Exercise 31.3 Assume that for each \(t\) that \(f_t\) is a continuous function of \(x\), and let \(t_k\to t\), \(x_k\to x\) be sequences such that for each \(k\in\NN\), the difference quotient \[\left|\frac{f_{t_k}(x_k)-f_t(x)}{t_k-t}\right|>k\] Show that replacing \(x_k\) with its limit \(x\) in the above gives an expression that tends to infinity in the limit: thus \(f_t\) is not differentiable at \(x\).
*Hint: add zero in a clever way to the top as \(0=f_{t_k}(x)-f_{t_k}(x)\) and re-arrange, using continuity of \(f_{t_k}\) as a function of \(x\).