Integration by substitution for Lebesgue integration

First show it is true for characteristic functions of closed intervals.

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).

(The label DCT means that I have implicitly used the dominated convergence theorem.)