Lebesgue integration by substitution

I read that, if $f\in L^1[c,d]$ is a Lebesgue summable function on $[a,b]$ and $g:[a,b]\to[c,d]$ is invertible and such that $g\in C^1[a,b]$ and $g^{-1}\in C^1[a,b]$, then $$\int_\limits{g([a,b])}f(x)\,d\mu_x=\int_\limits{[a,b]}f(g(t))|g'(t)|\,d\mu_t$$where $\mu$ is the linear Lebesgue measure.

I know that the function $F$ defined by $$F(x):=\int_\limits{[c,x]}f(\xi)\,d\mu_{\xi}$$is absolutely continuous, and that the derivative $\varphi$ of an absolutely continuous function $\Phi:[c,d]\to\mathbb{R}$, which exists almost everywhere on $[c,d]$, is such that $$\int_\limits{[c,d]}\varphi(\xi) \,d\mu_{\xi}=\Phi(d)-\Phi(c)$$but I cannot use these two facts alone to prove the desired result.

I do see, for ex. for a non-decreasing $g$, that $\frac{d}{dt}\int_\limits{[g(a),g(t)]}f(x)\,d\mu_x=F'(g(t))g'(t)$ exists and is equal to $f(g(t))g'(t)$ for almost every $g(t)$ (and therefore for almost every $t$, since I think that this implies that a homeomorphism like $g$ maps null measure sets to null measure sets), but I am not able to derived the desired identity from this. How can it be proved? I thank you any answerer very much!

Solutions Collecting From Web of "Lebesgue integration by substitution"

The integrand $f$ can (and should) essentially be ignored here.
It’s probably clearer to prove a more general substitution rule. If $g: X\to Y$ is a map between measure spaces and $\mu$ is a measure on $X$, then its image measure $g\mu$ is defined by $(g\mu)(B)=\mu(g^{-1}(B))$ (on a suitable $\sigma$-algebra on $Y$). This comes with a substitution rule
\int_Y f(y)\, d(g\mu)(y) = \int_X f(g(x))\, d\mu(x) ;
to prove this, start out by checking it for $f=\chi_B$ (this holds by the definition of $g\mu$) and then for general measurable $f\ge 0$ by approximating by simple functions, and then we have it for general $f\in L^1$.

Now back to your case: it suffices to show that Lebesgue measure on $Y=[c,d]$ is the image measure of $d\mu=|g’|\, dt$ on $X=[a,b]$, under the map $g$ (let’s call this measure $\nu$). This is obvious because it holds for intervals:
\nu(p,q) = \mu(g^{-1}(p,q)) = \int_{g^{-1}(p)}^{g^{-1}(q)} g'(t)\, dt = g(g^{-1}(p)) – g(g^{-1}(q)) = p-q,
as desired (here I assumed $g’>0$, but of course the calculation is analogous in the other case).