Suppose $f:[0,1] \to \mathbb{R}$ is bounded and measurable and $g:[-2,2]\to \mathbb{R}$ is Lebesgue integrable. We need to show that
$\lim_{h\to 0} \int_{0}^1 f(x)g(x+h) = \int_0^1 f(x)g(x) $.
Attempt 1:
We know that $|f(x)g(x+h)|\leq M |g(x+h)| $ where $g(x+h)$ is integrable on $[0,1]$ for small $h$. By DCT, we should have
$\lim_{h\to 0} \int f(x)g(x+h)= \int_0^1 \lim_{h\to 0} f(x)g(x+h)$
If only we knew $\lim_{h\to 0} f(x)g(x+h)=f(x)g(x)$ a.e we would be done.
This would indeed be true if $g$ was continuous. I don't know how to proceed from here except to do it for simple functions (which approximate $g$) and then try to argue it for $g$. In that argument also, I need to interchange two limits which I am not sure is valid or not.
Attempt 2:
Since $f$ is measurable, it is almost continuous (Luzin's theorem), we could "somehow" transfer the $h$ to $f$ instead of $g$ (commutativity of convolution) and then perhaps the result would be easier since we know that
$\lim_{h\to 0} f(x+h)g(x)=f(x)g(x)$ a.e
Any suggestions are welcome. Thanks and merry christmas.
Hint:
Use the fact that $C$ is dense in $L^1$ and
$$\left|\int_0^1 f(x) g(x+h) - \int_0^1f(x)g(x)\right| \leqslant M\int_0^1|g(x+h) - g(x)|$$
For any $\epsilon> 0$ there exists $g_\epsilon \in C([-2,2])$ such that
$$\int_0^1|g(x) - g_\epsilon(x)| < \epsilon.$$
and
$$\int_0^1|g(x+h) - g_\epsilon(x+h)| =\int_h^{1+h}|g(x) - g_\epsilon(x)| \\ = \int_h^{1}|g(x) - g_\epsilon(x)| + \int_1^{1+h} |g(x) - g_\epsilon(x)| \\ \epsilon + \int_1^{1+h} |g(x) - g_\epsilon(x)|. $$
Do you see why the last integral can be made small for $h$ sufficiently small?