# How to prove integration by substitution for Lebesgue measure?

The problem I'm having trouble answering is: Suppose $h$ is a differentiable function on $[a,b]$ with a continuous, positive derivative $h'(y)$ for all $x \in [a,b]$. For a measurable subset $\lambda\subset[a,b]$, show that $m(h(\lambda)) = \int_{\lambda}h'$. Then, use this to prove the Integration by substitution formula, namely that $$\int_a^bf(g(x))g'(x)dx = \int_{g(a)}^{g(b)}f(t)dt.$$

Does anyone have any suggestions on how I should go about solving this?

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I made some minor changes in formatting and chose a slightly more informative title ("measure and integration") is a bit too generic to be of any use in searches. :) If you don't like it you can always change it again. –  t.b. Mar 16 '12 at 23:20
You just need $F$ to be Absolutely continuous and increasing! –  checkmath Mar 17 '12 at 0:39
The idea of the title of a question would be that they reveal something about the content. "Proving measurability" has absolutely nothing to do with your question. Please do make an effort and be more specific. –  t.b. Mar 17 '12 at 21:46

$\bf Hint:$ If $h'(y)$ is always positive then $h$ is an increasing function. It follows that for any interval $I=(c,d)\subseteq [a,b], \ h((c,d))=(h(c),h(d))$. In this particular case $m(h(I))=m((h(c),h(d))=h(d)-h(c)=\int_{(c,d)}h'$. For the general case, use approximations to $\lambda$ by open sets.
@Patrick You are right $d-c$ should be $h(d)-h(c)$ and the result follows from the Fundamental Theorem of Calculus. –  azarel Mar 17 '12 at 15:46