# Differentiability under the integral sign of absolutely continuous $f(-,x)$

I read that

if

• $f:[a,b]\times X\mapsto\mathbb{R}$, where $X$ is a space endowed with measure$\mu_x$ and we define in $[a,b]\subset\mathbb{R}$ the usual Lebesgue linear measure $\mu_t$, is a $\mu_x\otimes\mu_t$-measurable function,
• for almost all $t\in[a,b]\quad$ $f(t,-)\in L^1(X,\mu_x)$, i.e. $x\mapsto f(t,x)$ is Lebesgue integrable on $X$,
• for almost all $x\in X\quad$ $f(-,x)$ (i.e. $t\mapsto f(t,x)$) is absolutely continuous,
• $\int_{[a,b]}\int_X\left|\frac{\partial}{\partial t}f(t,x)\right|d\mu_xd\mu_t<\infty$

then $$t\mapsto\int_Xf(t,x)d\mu_x$$is absolutely continuous and for almost all $t\in[a,b]$ $$\frac{d}{dt}\int_Xf(t,x)d\mu_x=\int_X\frac{\partial}{\partial t}f(t,x)d\mu_x$$

I see that, if $t\mapsto\int_Xf(t,x)d\mu_x$ is absolutely continuous on all $[a,b]$, and therefore differentiable almost everywhere on $[a,b]$, a generalisation of evaluation theorem to the derivative of an absolutely continuous function guarantees that, for all $\tau\in[a,b]$, $$\int_X\int_{[a,\tau]}\frac{\partial}{\partial t}f(t,x)d\mu_td\mu_x=\int_Xf(\tau,x)-f(a,x)d\mu_x.$$Fubini's theorem then allows to write$$\frac{d}{d\tau}\int_Xf(\tau,x)d\mu_x=\frac{d}{d\tau}\int_{[a,\tau]}\int_X\frac{\partial}{\partial t}f(t,x)d\mu_xd\mu_t$$and the right member is $\int_X\frac{\partial}{\partial t}f(t,x)d\mu_x$ almost everywhere on $[a,b]$, thanks to a generalisation of the fundamental theorem of calculus.

But how can we prove that $t\mapsto\int_Xf(t,x)d\mu_x$ is absolutely continuous on all $[a,b]$? If the third condition meant that$$\forall\varepsilon>0\,\exists\delta>0:\text{for any partition of }[a,b]\text{ into intervals }(a_k,b_k),k=1,\ldots,n\quad\sum_{k=1}^n(b_k-a_k)<\delta\Rightarrow\forall x\in X\quad\sum_{k=1}^n|f(b_k,x)-f(a_k,x)|<\varepsilon$$and if $\mu_x(X)<\infty$, I think that the absolute continuity would be quite trivial, but I do not know how to prove it without such conditions. Moreover, a thing that I find perplexing is that $f(-,x)$ is said to be Lebesgue summable for almost all $t$ and not necessarily all... I heartily thank any answerer!

• You find this $\delta$ for all $x\in X$ uniformly, right? How can you be sure that this is possible? – Friedrich Philipp Mar 4 '16 at 20:08
• @FriedrichPhilipp Thank you for noticing the error! Edited post. – Self-teaching worker Mar 4 '16 at 20:22
• $\partial f/\partial t$ is a (pointwise) limit, and pointwise limits of measurable functions are measurable. – Friedrich Philipp Mar 4 '16 at 20:28