Show that function introduced by varying upper bound in Lebesgue integral is weakly differentiable Let $I$ be an open interval, $p\in[1,\infty[$, $u\in L_p(I)$ and $x_0\in I$. Consider $v:=\int_{x_0}^\cdot u(\xi)d\xi$. I am asked to show that $v$ is weakly differentiable. 
So, I thought we could get $u\in L_1([\inf I+\varepsilon, b])$ and $x_0,x\in[I+\varepsilon, b]$ for any $x\in I$ for some $\varepsilon>0$ and some $b\in I$, so $v$ should be differentiable at all $x\in I$ (in the classical sense), and weak differentiability would follow trivially...
But, the solution uses change of order of integration to directly show the defining property of weakly differentiable functions. Why? Did I miss something?
 A: $v$ is absolutely continuous on $I$ and hence differentiable a.e. on $I$, it is not necessarily the case that $v$ is differentiable everywhere on $I$. Note that $u$ is only Lebesgue integrable, it is not necessarily belong to $C^{1}(I)$ so $v$ cannot be thought as the classical Riemann version of fundamental theorem of calculus which asserting that the anti-derivative is differentiable. 
So it is not necessarily that $v'(x)=u(x)$ everywhere. 
As @David C. Ullrich has noted, being differentiable a.e. in the classical sense still does not entail that being weakly differentiable. The reasoning is the following: Assume that $v'$ exists a.e., then for $\varphi\in C_{0}^{\infty}(a,b)$, 
\begin{align*}
\int_{a}^{b}v'(x)\varphi(x)dx=\color{red}{v(x)\varphi(x)\bigg|_{x=a}^{x=b}}-\int_{a}^{b}v(x)\varphi'(x)dx,
\end{align*}
the question is the red label one. Let us review how do we deduce the formula for integration by parts. For Lebesgue integrals version, one does
\begin{align*}
\int_{a}^{b}v'(x)\varphi(x)dx+\int_{a}^{b}v(x)\varphi'(x)dx&=\int_{a}^{b}[v'(x)\varphi(x)+v(x)\varphi'(x)]dx\\
&=\int_{a}^{b}(v(x)\varphi(x))'dx\\
&=\color{red}{v(b)\varphi(b)-v(a)\varphi(a)}.
\end{align*}
If $v$ is differenitable everywhere and $v'$ is bounded, then $v$ is absolutely continuous. Since $\varphi$ is twice differentiable and has compactly supported, $\varphi$ is absolutely continuous, so the product $v\varphi$ is absolutely continuous, and the red label goes through.
If $v$ is merely differentiable everywhere but not absolutely continuous, the product $v\varphi$ is not necessarily absolutely continuous, so the red label may not go through.
Of course, in this question, $v$ is an anti-derivative of $u$, and we know that anti-derivative is absolutely continuous, so the product $v\varphi$ is absolutely continuous, and the red label goes through.   
In the classical mathematical analysis (Riemann integration), the formula for integration by parts only assumes that both $f$ and $g$ being differentiable with $f'$ and $g'$ being Riemann integrable. Note that Riemann integrable implies boundedness (it depends which definition one adopts, it may be required as a part of the definition of being Riemann integrable). So the assumption here entails that the product $fg$ is absolutely continuous, and it agrees with the Lebesgue integrals version.
A: The notation in the definition of $v$ could be clearer. More important, "weakly differenntiable" means various different things in various contexts - you should include a definition.
Here's a special case:


Suppose $f\in L^1(\Bbb R)$ and define $F(x)=\int_{-\infty}^x f(t)\,dt$. Then $F'=f$ weakly, in the sense that if $\phi\in C^1_c(\Bbb R)$ then $\int F\phi'=-\int f\phi$.


Note Yes, we know that $F'=f$ almost everywhere. Best to forget about that! Because it really doesn't help in showing that $F'=f$ in the weak sense defined above. In fact  the fact that $F'=f$ almost everywhere is much deeper  than what we're proving here.
Example showing that $F'=f$ almost everywhere does not imply weak differentiability: Let $F$ be the Cantor-Lebesgue function, aka Devil's staircase. Then $F'=0$ almost everywhere. But it is not true that $F'=0$ weakly as above, because (exercise) this would imply that $F$ was constant.
Proof: Fubini shows that $$\int F(x)\phi'(x)=\int_{-\infty}^\infty \phi'(x)\int_{-\infty}^x f(t)\,dt\,dx=\int_{-\infty}^\infty f(t)\int_t^\infty\phi'(x)\,dx\,dt=-\int f(t)\phi(t)\,dt.$$


Interesting Note In fact if $f$ and $F$ are as above then $F'=f$ (strongly) "in $L^1$", in the following sense: If we say that $\tau_h F(t)=F(t+h)$ then $$\left|\left|f-\frac{\tau_h F-F}{h}\right|\right|_1\to0\quad(h\to0).$$


Proof: Exercise. (Once again, don't start with $F'=f$ almost everywehere; it won't help.)
