# Integration by substitution for Lebesgue integration

This is an excersise in Shiryaev's Probability on page 211

Starting from simple functions and using the theorem on taking limits under the Lebesgue integral sign, prove the following result on integration by substitution.

Let $h=h(y)$ be a nondecreasing continuously differentiable function on $[a, b]$, and let $f(x)$ be (Lebesgue) integrable on $[h(a), h(b)]$. Then the function $f(h(y))h'(y)$ is integrable on $[a, b]$ and $$\int_{h(a)}^{h(b)}f(x)dx=\int_{a}^{b}f(h(y))h'(y)dy$$

I don't how to check it is right for the simple functions. Thank you!

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Perhaps you can reduce from simple functions to characteristic functions of measurable sets, to char. fcns of $G_{\delta}$ sets, to char. fcns. of open sets, to char. fcns. of open intervals. – Jonas Meyer Nov 30 '12 at 5:07
@JonasMeyer Yes. I have tried to prove for the characteristic funcions of measurable sets. And I reduce it to prove $m(f(A))=\int_A f'dx$ for $A\in \mathcal{B}([a, b])$. For $A$ is interval, it is obvious. Then I need to prove the set $$E=\{A\in\mathcal{B}([a, b]):m(f(A))=\int_A f'dx\}$$ is a $\sigma$-algebra. I first tried to prove the union of countable disjoint sets ${A_n}\in E$ also in E. But there is a subtle thing that ${f(A_n)}$ need not be disjoint. Then I think it can be proved by the monotone class theorem. – Danielsen Nov 30 '12 at 5:58

Let $[c,d] \subset [h(a),h(b)]$. Then we have $\int_{h(a)}^{h(b)} 1_{[c,d]} = d-c$. Now consider $\int_a^b 1_{[c,d]}(h(y)) h'(y) dy$. Let $\gamma = \inf h^{-1}\{c\}$, $\delta = \sup h^{-1}\{d\}$. Then $\int_a^b 1_{[c,d]}(h(y)) h'(y) dy = \int_a^b 1_{[\gamma,\delta]}(y) h'(y) dy = \int_\gamma^\delta h'(y) dy = h(\delta)-h(\gamma) = d-c$.
Hence the formula is true for characteristic functions of closed intervals, and by linearity, it holds for sums of such functions. Since $[c,d] = \{c\} \cup (c,d) \cup \{d\}$ (or equivalently, $1_{[c,d]} = 1_{\{c\}}+1_{(c,d)}+1_{\{d\}}$), it follows that it is true for characteristic functions of open intervals. Since open sets are the (at most) countable disjoint union of open intervals, it is true for open sets (DCT), and hence for closed sets (since $1_C = 1-1_{C^C}$).
The usual regularity argument shows that it holds for $F_\sigma$ (DCT) and $G_\delta$ (DCT) sets, and hence for any arbitrary measurable set (DCT). For this it follows that it is true for simple functions, and the general results follows from this (DCT).