Composing Integrability

In this question, we compose certain functions to preserve integrability.

Problem 1. Let $f : [a, b] \to \mathbb R$ be an integrable function, and suppose $m \leq f \leq M$ . Let $\phi : [m, M] \to \mathbb R$ be continuous. Prove that $\phi \circ f$ is integrable.

Solution. Fix $\epsilon > 0$ . Since $\phi$ is continuous, it is uniformly continuous. Thus, for any $k_1 > 0$ , there exists $\delta > 0$ such that for any $u,v \in [m,M]$ ,

$|u-v| < \delta \quad \Rightarrow \quad |\phi(u) - \phi(v)| < k_1 \cdot \epsilon.$

Since $f$ is integrable, for any $k_2 > 0$ , there exists a partition $P \subseteq [a, b]$ such that

$\displaystyle \sum_{i = 1}^{|P|} (M_i(f,P) - m_i(f,P))\Delta x_i = U(f, P) - L(f, P) < k_2 \cdot \epsilon.$

We need to treat the values $m_i(f,P)$ , $M_i(f, P)$ as inputs to $\phi$ . We observe that if $M_i(f,P) - m_i(f,P) < \delta$ , then for any $u,v \in [m_i(f, P), M_i(f, P)]$ , the continuity of $\phi$ yields

$|\phi(u) - \phi(v)| < k_1 \cdot \epsilon.$

Therefore, we will define the index sets

$\begin{aligned} A &:= \{i : M_i(f,P) - m_i(f,P) < \delta\}, \\ B &:= \{i : M_i(f,P) - m_i(f,P) \geq \delta\}. \end{aligned}$

For each $i$ ,

$\begin{aligned} M_i(\phi \circ f, P) - m_i(\phi \circ f, P) &= \sup_{u, v \in [x_{i-1}, x_i] } (\phi(u) - \phi(v)). \end{aligned}$

For $i \in A$ ,

$\displaystyle M_i(\phi \circ f, P) - m_i(\phi \circ f, P) \leq k_1 \cdot \epsilon.$

For $i \in B$ , since $\phi$ is bounded, suppose $|\phi| \leq K$ . Then

$\displaystyle M_i(\phi \circ f, P) - m_i(\phi \circ f, P) \leq 2K.$

Therefore,

$\displaystyle \begin{aligned} U(\phi \circ f, P) - L(\phi \circ f, P) &< \sum_{i \in A} (k_1 \cdot \epsilon) \cdot \Delta x_i + \sum_{i \in B} 2K \cdot \Delta x_i \\ &= (k_1 \cdot \epsilon) \cdot \sum_{i \in A} \Delta x_i + 2K \cdot \sum_{i \in B} \Delta x_i \\ &\leq (k_1 \cdot \epsilon) \cdot (b-a) + 2K \cdot \sum_{i \in B} \Delta x_i. \end{aligned}$

We have run into a problem. The first term can be controlled pretty well, but not the second. The key idea is to take advantage of the integrability of $f$ . Since each “height” is $\geq \delta$ and the region under $f$ has finite area, the total “width” needs to be tiny.

To synthesise this observation, we make the estimate

$\displaystyle \begin{aligned} k_2 \cdot \epsilon &> \sum_{i \in B} (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &\geq \sum_{i \in B} \delta \cdot \Delta x_i \\ &= \delta \cdot \sum_{i \in B} \Delta x_i. \end{aligned}$

This yields the estimate

$\displaystyle \sum_{i \in B} \Delta x_i < \frac{k_2 \cdot \epsilon}{\delta}.$

Therefore,

$\displaystyle \begin{aligned} U(\phi \circ f, P) - L(\phi \circ f, P) &\leq (k_1 \cdot \epsilon) \cdot (b-a) + 2K \cdot \sum_{i \in B} \Delta x_i \\ &<(k_1 \cdot \epsilon) \cdot (b-a) + 2K \cdot \frac{k_2 \cdot \epsilon}{\delta} \\ &= \left(k_1 \cdot (b-a) + 2K \cdot \frac{k_2}{\delta}\right) \cdot \epsilon. \end{aligned}$

Taking care of the dependencies, setting $k_1 = 1/(2(b-a))$ , $k_2 = \delta/(4K)$ yields the desired result.

Problem 2. Construct an integrable function $f : [a, b] \to \mathbb R$ be an integrable function with $m \leq f \leq M$ and an integrable function $\phi : [m, M] \to \mathbb R$ such that $\phi \circ f$ is not integrable.

Solution. Letting $f : [0, 1] \to \mathbb R$ be the popcorn function and defining $\phi : [0, 1] \to \mathbb R$ by $\phi(x) = 0$ if $x = 0$ and $\phi(x) = 1$ otherwise. Both $f, \phi$ are integrable. Then $\phi \circ f = \mathbb I_{\mathbb Q} : [0, 1] \to \mathbb R$ is a bounded function known as the Dirichlet function. We claim that this function is not Riemann-integrable.

Let $\epsilon > 0$ be suitably chosen and $P = \{x_0,x_1,\dots,x_n\}$ be any partition of $[0, 1]$ . For each $i$ , since $\mathbb Q \cap [x_{i-1}, x_i]$ and $[x_{i-1}, x_i]\backslash \mathbb Q$ are both dense in $[x_{i-1}, x_i]$ , there exist $r \in \mathbb Q \cap [x_{i-1}, x_i]$ , $s \in [x_{i-1}, x_i]\backslash \mathbb Q$ , yielding

$\begin{aligned} 0 &\leq m_i(f, P) \leq f(s) = 0,\\ 1 &= f(r) \leq M_i(f, P) \leq 1.\end{aligned}$

so that $m_i(f, P) = 0$ , $M_i(f, P) = 1$ . Therefore,

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i =1}^n (M_i(f, P) - m_i(f,P))\Delta x_i \\ &= \sum_{i =1}^n (1 - 0)\Delta x_i \\ &= (1 - 0)\cdot \sum_{i =1}^n \Delta x_i \\ &= (1 - 0) \cdot 1 = 1. \end{aligned}$

Setting $\epsilon = 1/2 < 1$ yields the result that the Dirichlet function is not integrable.

—Joel Kindiak, 21 Jan 25, 0033H

KindiakMath

Composing Integrability

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Composing Integrability

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