$F(t) = \int_0^1 f(x+\psi(t)) dx$, $\psi \in C^1([0,1])$ and $\psi'>0$ imply $F$ differentiable a.e. I am trying to solve the following problem: 
Let $f\in L^1([0,2])$, let $\psi: [0,1]\to[0,1]$ be a function, and let $F: [0,1]\to \mathbb{R}$ be defined by 
$$
F(t) = \int_0^1 f(x+\psi(t)) dx \quad \text{ for every } t \in [0,1].
$$


*

*Prove that, if $\psi$ is continuous, then $F$ is continuous.

*Prove that, if $\psi \in C^1([0,1])$ and $\psi'(t)>0$ for every $t \in [0,1]$, then $F$ is differentiable for almost every $t \in [0,1]$
My attempt: 
For point (1) I changed variable, setting $y = x + \psi(t)$, $dy = dx$ to get 
$$
F(t) = \int_{\psi(t)}^{1+\psi(t)} f(y) dy = \int_0^2 f(y) \chi_{(\psi(t), 1+\psi(t))}(y) dy.
$$
Then I concluded that $F$ is continuous since, for every sequence $\{t_n\}_n$ such that $t_n \to t_0$, we have $F(t_n) \to F(t_0)$ by the dominated convergence theorem. 
For point (2), since I have $f \in L^1$, I thought of using the well-known result: 

If $g \in L^1(a,b)$ and $G(t) = \int_a^t g(x)dx$ for every $t \in [a,b]$, then $G$ is absolutely continuous on $[a,b]$ and there exists $G'(t) = g(t) $ for almost every $t \in [a,b]$.

With this in mind, I wrote $F(t)$ as 
$$
F(t) = \int_0^{1+\psi(t)} f(y) dy - \int_0^{\psi(t)} f(y) dy.
$$
However, I notice that I do not have $t$ as upper bound of integration, but a function of $t$. 
If $f$ had more regularity, I could simply compute 
$$
F'(t) = \psi'(t) (f(1+\psi(t))-f(\psi(t))), 
$$
but I feel that this requires more justification in this context (and I did not use the fact that $\psi' >0 $, so it must be wrong). 
Could you please give me some help on how to prove point (2)? 
P.s. This problem is taken from a past entrance exam to a PhD in Mathematical Analysis. If you recognize that this is from some book, or if you have a source of similar problems, please tell me.
 A: Let $G(t)= \int_0^t f(x)\,dx.$ Then $G'(t)$ exists for a.e. $t \in [0,2].$  Note that
$$F(t)= G(1+\psi(t))- G(\psi(t)).$$
By the chain rule, we're done if we show both $G'(1+\psi(t)),G'(\psi(t))$ exist for a.e. $t\in [0,1].$
Let $E$ be the set of $t\in [0,2]$ such that $G'(t)$ fails to exist. We know $m(E)=0.$ Now $G'(\psi(t))$ exists if $\psi(t)\notin E,$ i.e., if $t\notin\psi^{-1}(E).$ But here's the thing: $\psi^{-1}$ is $C^1$ and $C^1$ functions preserve sets of measure $0.$ Thus $\psi^{-1}(E)$ has measure $0.$ Hence $G'(\psi(t))$ exists for a.e. $t.$
The same argument applies to $G'(1+\psi(t)).$ Thus both $G'(1+\psi(t)),G'(\psi(t))$ exist for a.e. $t\in [0,1]$ as desired, and we're done.
A: Concerning (2): the accepted answer actually uses the rather strong assumptions $\psi \in C^1([0,1])$ and $\psi'(t)>0$ for every $t \in [0,1]$, and it proves a stronger result, $$F'(t) = \psi'(t) (f(1+\psi(t))-f(\psi(t)))\quad\mbox{a.e.}$$
If we really are interested just in "$F$ is differentiable a.e.", the (weaker) assumption "$\psi$ is monotone increasing" is sufficient.
Proof: We can write $f=g-h$ with non-negative integrable functions $g$ and $h$, e.g. $g=|f|$ and $h=|f|-f.$ With $G(t)=\int^t_0g(x)\,dx$ and $H(t)=\int^t_0h(x)\,dx,$ we have
$$F(t)=(G(1+\psi(t))+H(\psi(t)))-(G(\psi(t))+H(1+\psi(t))),$$ i.e. the difference of two monotone functions. But according to Lebesgue's Theorem (http://mathonline.wikidot.com/lebesgue-s-theorem-for-the-differentiability-of-monotone-fun), monotone functions are differentiable a.e. 
