Differentiability for an integral function

Question

I can't conclude the following exercise:

Let $f in L^{1}[0,2]$ , $psi:[0,1] rightarrow [0,1]$ a function and $F[0,1] rightarrow mathbb{R}$ defined by $$F(t)= int_0^1 f(x+psi(t))dx$$ for every $ t in [0,1]$.

Show that if $psi$ is continuous then $F$ is continuous.
Show that if $psi in C^{1}$ and $psi'(t)>0$ for every $t in [0,1]$, then $F$ is differentiable for almost every $ t in [0,1]$.

Let $s<t$: I consider the difference

$$|F(t)-F(s)| = left| int_0^1Bigl[f(x+psi(t)) - f(x+psi(s)) Bigr] dx right| leq int_0^1 left| f(x+ psi(t)) - f(x+psi(s)) right|dx$$
Taking the limit as $s rightarrow t^{-}$ and using the fact that $f in L^{1}$, I can pass the limit inside the sign of integral in the rhs, and I obtain the continuity.

Here if $f$ would be differentiable, everything would work. Since it is only $L^1$, this function can't be differentiated, so I was thinking to split the variables in $f(x+psi(t))$, but I don't know how.

Any help/answer/hint is highly appreciated

Sangchul Lee · Accepted Answer

As pointed out by @Kavi Rama Murthy, your proof of continuity is not correct as one cannot deduce the a.e.-pointwise convergence of the integrand.
(a) For the continuity, recast the integral as
$$ F(t) = int_{0}^{2} f(x)mathbf{1}_{[psi(t),psi(t)+1]}(x) , mathrm{d}x $$
and then note that now one can apply the dominated convergence theorem.
(b) The above representation also allows to solve the second part. To this end, let
$$ E = { u in [0, 1] : text{both $u$ and $u+1$ are Lebesgue points of $f$} }. $$
If $Z$ denotes the complement of the set of Lebesgue points of $f$ in $[0, 2]$, then $Z$ has measure zero by the Lebesgue differentiation theorem. Moreover,
$$[0,1]setminus E subseteq Z cup (Z-1), $$
and so, $E$ has full measure in $[0,1]$. Now by the assumption, $psi$ has a $C^1$ inverse, and so, $psi^{-1}(E)$ also has full measure in $[0,1]$. Finally, if $t in psi^{-1}(E)$, then both $psi(t)$ and $psi(t)+1$ are Lebesgue points of $f$, and so,
$$ frac{F(t+h) - F(t)}{h} = frac{1}{h} int_{psi(t)+1}^{psi(t+h)+1} f(x) , mathrm{d}x - frac{1}{h} int_{psi(t)}^{psi(t+h)} f(x) , mathrm{d}x $$
converges as $h to 0$. This completes the proof.

Differentiability for an integral function

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