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Section 10.4 The set of Riemann integrable functions

Note: 1 lecture

Subsection 10.4.1 Oscillation and continuity

Consider \(D \subset \R^n\) and \(f \colon D \to \R\text{.}\) Instead of just saying that \(f\) is or is not continuous at a point \(x \in D\text{,}\) we want to quantify how discontinuous \(f\) is at \(x\text{.}\) For every \(\delta > 0\text{,}\) define the oscillation of \(f\) on the \(\delta\)-ball in subspace topology, \(B_D(x,\delta) = B_{\R^n}(x,\delta) \cap D\text{,}\) as

\begin{equation*} o(f,x,\delta) := {\sup_{y \in B_D(x,\delta)} f(y)} - {\inf_{y \in B_D(x,\delta)} f(y)} = \sup_{y_1,y_2 \in B_D(x,\delta)} \bigl(f(y_1)-f(y_2)\bigr) . \end{equation*}

That is, \(o(f,x,\delta)\) is the length of the smallest interval that contains the image \(f\bigl(B_D(x,\delta)\bigr)\text{.}\) For unbounded functions, the oscillation could be \(\infty\text{,}\) although we only need to worry about bounded functions. Clearly \(o(f,x,\delta) \geq 0\) and \(o(f,x,\delta) \leq o(f,x,\delta')\) whenever \(\delta < \delta'\text{.}\) Therefore, the limit as \(\delta \to 0\) from the right exists, and we define the oscillation of \(f\) at \(x\) as

\begin{equation*} o(f,x) := \lim_{\delta \to 0^+} o(f,x,\delta) = \inf_{\delta > 0} o(f,x,\delta) . \end{equation*}

Proof.

First suppose that \(f\) is continuous at \(x \in D\text{.}\) Given any \(\epsilon > 0\text{,}\) there exists a \(\delta > 0\) such that for \(y \in B_D(x,\delta)\text{,}\) we have \(\sabs{f(x)-f(y)} < \epsilon\text{.}\) Therefore, if \(y_1,y_2 \in B_D(x,\delta)\text{,}\) then

\begin{equation*} f(y_1)-f(y_2) = \bigl(f(y_1)-f(x)\bigr)-\bigl(f(y_2)-f(x)\bigr) < \epsilon + \epsilon = 2 \epsilon . \end{equation*}

We take the supremum over \(y_1\) and \(y_2\)

\begin{equation*} o(f,x,\delta) = \sup_{y_1,y_2 \in B_D(x,\delta)} \bigl(f(y_1)-f(y_2)\bigr) \leq 2 \epsilon . \end{equation*}

As \(o(x,f) \leq o(f,x,\delta) \leq 2\epsilon\text{,}\) and \(\epsilon > 0\) was arbitrary, \(o(x,f) = 0\text{.}\)

On the other hand suppose \(o(x,f) = 0\text{.}\) Given any \(\epsilon > 0\text{,}\) find a \(\delta > 0\) such that \(o(f,x,\delta) < \epsilon\text{.}\) If \(y \in B_D(x,\delta)\text{,}\) then

\begin{equation*} \sabs{f(x)-f(y)} \leq \sup_{y_1,y_2 \in B_D(x,\delta)} \bigl(f(y_1)-f(y_2)\bigr) = o(f,x,\delta) < \epsilon. \qedhere \end{equation*}

Proof.

Equivalently, we want to show that \(G := \bigl\{ x \in D : o(f,x) < \epsilon \bigr\}\) is open in the subspace topology. Consider \(x \in G\text{.}\) As \(\inf_{\delta > 0} o(f,x,\delta) < \epsilon\text{,}\) find a \(\delta > 0\) such that

\begin{equation*} o(f,x,\delta) < \epsilon \end{equation*}

Take any \(\xi \in B_D(x,\nicefrac{\delta}{2})\text{.}\) Notice that \(B_D(\xi,\nicefrac{\delta}{2}) \subset B_D(x,\delta)\text{.}\) Therefore,

\begin{equation*} o(f,\xi,\nicefrac{\delta}{2}) = \sup_{y_1,y_2 \in B_D(\xi,\nicefrac{\delta}{2})} \bigl(f(y_1)-f(y_2)\bigr) \leq \sup_{y_1,y_2 \in B_D(x,\delta)} \bigl(f(y_1)-f(y_2)\bigr) = o(f,x,\delta) < \epsilon . \end{equation*}

So \(o(f,\xi) < \epsilon\) as well. As this is true for all \(\xi \in B_D(x,\nicefrac{\delta}{2})\text{,}\) we get that \(G\) is open in the subspace topology and \(D \setminus G\) is closed as claimed.

Subsection 10.4.2 The set of Riemann integrable functions

We have seen that continuous functions are Riemann integrable, but we also know that certain kinds of discontinuities are allowed. It turns out that as long as the discontinuities happen on a set of measure zero, the function is integrable, and vice versa.

Proof.

Let \(S \subset R\) be the set of discontinuities of \(f\text{,}\) that is, \(S = \bigl\{ x \in R : o(f,x) > 0 \bigr\}\text{.}\) Suppose \(S\) is a measure zero set: \(m^*(S) = 0\text{.}\) The trick to proving that \(f\) is integrable is to isolate the bad set into a small set of subrectangles of a partition. A partition has finitely many subrectangles, so we need compactness. If \(S\) were closed, then it would be compact and we could cover it by finitely many small rectangles. Unfortunately, \(S\) itself is not closed in general, but the following set is.

Given \(\epsilon > 0\text{,}\) define

\begin{equation*} S_\epsilon := \bigl\{ x \in R : o(f,x) \geq \epsilon \bigr\} . \end{equation*}

By Proposition 10.4.2, \(S_\epsilon\) is closed and as it is a subset of \(R\text{,}\) which is bounded, \(S_\epsilon\) is compact. Furthermore, \(S_\epsilon \subset S\) and \(S\) is of measure zero, so \(S_\epsilon\) is of measure zero. Via Proposition 10.3.7, finitely many open rectangles \(O_1,O_2,\ldots,O_k\) cover \(S_\epsilon\) and \(\sum V(O_j) < \epsilon\text{.}\)

The set \(T := R \setminus ( O_1 \cup \cdots \cup O_k )\) is closed, bounded, and so compact. As \(o(f,x) < \epsilon\) for all \(x \in T\text{,}\) for each \(x \in T\text{,}\) there is a \(\delta > 0\) such that \(o(f,x,\delta) < \epsilon\text{,}\) so there exists a small closed rectangle \(T_x \subset B(x,\delta)\) with \(x\) in the interior of \(T_x\text{,}\) such that

\begin{equation*} \sup_{y\in T_x} f(y) - \inf_{y\in T_x} f(y) < \epsilon. \end{equation*}

The interiors of the rectangles \(T_x\) cover \(T\text{.}\) As \(T\) is compact, finitely many such rectangles \(T_1, T_2, \ldots, T_m\) cover \(T\text{.}\) Take the rectangles \(T_1,T_2,\ldots,T_m\) and \(O_1,O_2,\ldots,O_k\) and construct a partition out of their endpoints. That is, construct a partition \(P\) of \(R\) with subrectangles \(R_1,R_2,\ldots,R_p\) such that every \(R_j\) is contained in \(T_\ell\) for some \(\ell\) or the closure of \(O_\ell\) for some \(\ell\text{.}\) Order the rectangles so that \(R_1,R_2,\ldots,R_q\) are those that are contained in some \(T_\ell\text{,}\) and \(R_{q+1},R_{q+2},\ldots,R_{p}\) are the rest. So

\begin{equation*} \sum_{j=1}^q V(R_j) \leq V(R) \qquad \text{and} \qquad \sum_{j=q+1}^p V(R_j) \leq \sum_{\ell=1}^k V(O_\ell) < \epsilon . \end{equation*}

Let \(m_j\) and \(M_j\) be the inf and sup of \(f\) over \(R_j\) as before. If \(R_j \subset T_\ell\) for some \(\ell\text{,}\) then \(M_j-m_j < 2 \epsilon\text{.}\) Let \(B \in \R\) be such that \(\sabs{f(x)} \leq B\) for all \(x \in R\text{,}\) so \(M_j-m_j \leq 2B\) over all rectangles. Then

\begin{equation*} \begin{split} U(P,f)-L(P,f) & = \sum_{j=1}^p (M_j-m_j) V(R_j) \\ & = \left( \sum_{j=1}^q (M_j-m_j) V(R_j) \right) + \left( \sum_{j=q+1}^p (M_j-m_j) V(R_j) \right) \\ & < \left( \sum_{j=1}^q 2\epsilon\, V(R_j) \right) + \left( \sum_{j=q+1}^p 2 B\, V(R_j) \right) \\ & < 2 \epsilon\, V(R) + 2B \epsilon = \epsilon \bigl(2\, V(R)+2B\bigr) . \end{split} \end{equation*}

We can make the right-hand side as small as we want, and hence \(f\) is integrable.

For the other direction, suppose \(f\) is Riemann integrable on \(R\text{.}\) Let \(S\) be the set of discontinuities again. Consider the sequence of sets

\begin{equation*} S_{1/k} = \bigl\{ x \in R : o(f,x) \geq \nicefrac{1}{k} \bigr\}. \end{equation*}

Fix a \(k \in \N\text{.}\) Given an \(\epsilon > 0\text{,}\) find a partition \(P\) with subrectangles \(R_1,R_2,\ldots,R_p\) such that

\begin{equation*} U(P,f)-L(P,f) = \sum_{j=1}^p (M_j-m_j) V(R_j) < \epsilon \end{equation*}

Suppose \(R_1,R_2,\ldots,R_p\) are ordered so that the interiors of \(R_1,R_2,\ldots,R_{q}\) intersect \(S_{1/k}\text{,}\) while the interiors of \(R_{q+1},R_{q+2},\ldots,R_p\) are disjoint from \(S_{1/k}\text{.}\) If \(x \in R_j \cap S_{1/k}\) and \(x\) is in the interior of \(R_j\text{,}\) then \(o(f,x) \geq \nicefrac{1}{k}\text{.}\) As sufficiently small \(\delta\)-balls are completely inside \(R_j\) and \(o(f,x,\delta) \geq o(f,x) \geq \nicefrac{1}{k}\text{,}\) we get \(M_j-m_j \geq \nicefrac{1}{k}\text{.}\) Then

\begin{equation*} \epsilon > \sum_{j=1}^p (M_j-m_j) V(R_j) \geq \sum_{j=1}^q (M_j-m_j) V(R_j) \geq \frac{1}{k} \sum_{j=1}^q V(R_j) \end{equation*}

In other words, \(\sum_{j=1}^q V(R_j) < k \epsilon\text{.}\) Let \(G\) be the set of all boundaries of all the subrectangles of \(P\text{.}\) The set \(G\) is of measure zero (as it can be covered by finitely many sets from Example 10.3.5). Let \(R_j^\circ\) denote the interior of \(R_j\text{,}\) then

\begin{equation*} S_{1/k} \subset R_1^\circ \cup R_2^\circ \cup \cdots \cup R_q^\circ \cup G . \end{equation*}

As \(G\) can be covered by open rectangles arbitrarily small volume, \(S_{1/k}\) must be of measure zero. As

\begin{equation*} S = \bigcup_{k=1}^\infty S_{1/k} \end{equation*}

and a countable union of measure zero sets is of measure zero, \(S\) is of measure zero.

The proof is contained in the exercises.

Subsection 10.4.3 Exercises

Exercise 10.4.1.

Suppose \(f \colon (a,b) \times (c,d) \to \R\) is a bounded continuous function. Show that the integral of \(f\) over \(R = [a,b] \times [c,d]\) makes sense and is uniquely defined. That is, set \(f\) to be anything on the boundary of \(R\) and compute the integral, showing that the values on the boundary are irrelevant.

Exercise 10.4.2.

Suppose \(R \subset \R^n\) is a closed rectangle. Show that \(\sR(R)\text{,}\) the set of Riemann integrable functions, is an algebra. That is, show that if \(f,g \in \sR(R)\) and \(a \in \R\text{,}\) then \(af \in \sR(R)\text{,}\) \(f+g \in \sR(R)\text{,}\) and \(fg \in \sR(R)\text{.}\)

Exercise 10.4.3.

Suppose \(R \subset \R^n\) is a closed rectangle and \(f \colon R \to \R\) is a bounded function which is zero except on a closed set \(E \subset R\) of measure zero. Show that \(\int_R f\) exists and compute it.

Exercise 10.4.4.

Suppose \(R \subset \R^n\) is a closed rectangle and \(f \colon R \to \R\) and \(g \colon R \to \R\) are two Riemann integrable functions. Suppose \(f = g\) except for a closed set \(E \subset R\) of measure zero. Show that \(\int_R f = \int_R g\text{.}\)

Exercise 10.4.5.

Suppose \(R \subset \R^n\) is a closed rectangle and \(f \colon R \to \R\) is a bounded function.

  1. Suppose there exists a closed set \(E \subset R\) of measure zero such that \(f|_{R\setminus E}\) is continuous. Then \(f \in \sR(R)\text{.}\)

  2. Find an example where \(E \subset R\) is a set of measure zero (not closed) such that \(f|_{R\setminus E}\) is continuous and \(f \not\in \sR(R)\text{.}\)

Exercise 10.4.6.

Suppose \(R \subset \R^n\) is a closed rectangle and \(f \colon R \to \R\) and \(g \colon R \to \R\) are Riemann integrable. Show that

\begin{equation*} \varphi(x) := \max \bigl\{ f(x) , g(x) \bigr\} , \qquad \psi(x) := \min \bigl\{ f(x) , g(x) \bigr\} , \end{equation*}

are Riemann integrable.

Exercise 10.4.7.

Suppose \(R \subset \R^n\) is a closed rectangle and \(f \colon R \to \R\) is Riemann integrable. Show that \(\sabs{f}\) is Riemann integrable. Hint: Define \(f_+(x) := \max \bigl\{ f(x) , 0 \bigr\}\) and \(f_-(x) := \max \bigl\{ -f(x) , 0 \bigr\}\text{,}\) and then write \(\sabs{f}\) in terms of \(f_+\) and \(f_-\text{.}\)

Exercise 10.4.8.

  1. Suppose \(R \subset \R^n\) and \(R' \subset \R^n\) are closed rectangles, \(U \subset \R^n\) and \(U' \subset \R^n\) are open sets such that \(R \subset U\) and \(R' \subset U'\text{,}\) \(g \colon U \to U'\) is continuously differentiable, bijective, \(g^{-1}\) is continuously differentiable, \(g(R) \subset R'\text{,}\) and \(f \in \sR(R')\text{,}\) then the composition \(f \circ g\) is Riemann integrable on \(R\text{.}\)

  2. Find a counterexample when \(g\) is not one-to-one. Hint: Try \(g(x,y) := (x,0)\) and \(R=R'=[0,1] \times [0,1]\text{.}\)

Exercise 10.4.9.

Suppose \(f \colon [0,1]^2 \to \R\) is defined by

\begin{equation*} f(x,y) := \begin{cases} \frac{1}{kq} & \text{if } x,y \in \Q \text{ and } x = \frac{\ell}{k} \text{ and } y=\frac{p}{q} \text{ in lowest terms,} \\ 0 & \text{else.} \end{cases} \end{equation*}

Show that \(f \in \sR\bigl({[0,1]}^2\bigr)\text{.}\)

Exercise 10.4.10.

Compute the oscillation \(o\bigl(f,(x,y)\bigr)\) for all \((x,y) \in \R^2\) for the function

\begin{equation*} f(x,y) := \begin{cases} \frac{xy}{x^2+y^2} & \text{if } (x,y) \not= (0,0), \\ 0 & \text{if } (x,y) = (0,0) . \end{cases} \end{equation*}

Exercise 10.4.11.

Consider the popcorn function \(f \colon [0,1] \to \R\text{,}\)

\begin{equation*} f(x) := \begin{cases} \frac{1}{q} & \text{if } x \in \Q \text{ and } x = \frac{p}{q} \text{ in lowest terms,} \\ 0 & \text{else.} \end{cases} \end{equation*}

Compute \(o(f,x)\) for all \(x \in [0,1]\text{.}\)

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