Limit of Bounded measurable function

Question

I got a problem stated as below.

Problem Suppose that $f$ is an integrable function and $g$ is a bounded measurable function. Then prove that $$ \lim_{t\rightarrow 0} \int f(x)\left[ g(x+t)-g(x) \right]dx =0 $$

At first moment, I came up dominated convergence theorem since $$|f(x)\left[g(x+t)-g(x) \right]|\leq 2M|f(x)| \hspace{3mm}\forall t .$$ But I could not prove that $$\lim_{t\rightarrow 0} [g(x+t)-g(x)]=0 \hspace{4mm}a.e.$$ For now, I cannot come up with another way to approach it. Should I change the way to prove?

Answer 1

It is a standard fact that for any integrable function $f$, $\int |f(y+t)-f(y)|\, dy \to 0$ as $t \to 0$. [ Rudin's RCA has a proof]. Write the given integral as $\int [f(y-t)-f(y)] g(y)\, dy$ (by the substitution $y=x+t$) and pull out the sup of $|g|$]. This completes the proof. Some details about rewriting the given integral: $$\int f(x)[g(x+t)-g(x)]\, dx=\int f(x)g(x+t)\, dx-\int f(x)g(x)\, dx=\int f(y-t)g(y)\, dy -\int f(x)g(x)\, dx$$ by substitution. Hence $\int f(x)[g(x+t)-g(x)]\, dx= \int [f(y-t)-f(y)]g(y)\, dy$, so $|\int [f(y-t)-f(y)]g(y)\, dy| \leq M \int |f(y-t)-f(y)|\, dy$ where $M=\sup |g(x)|$.

Answer 2

Hint 1: Try using Egorov's Theorem. Approximate the domain of integration by a compact set $K$. There is an $\epsilon > 0$ such outside a set of measure $\epsilon$ on $K$, the limit is uniform.

Hint 2: You could also change variables, to interchange $f$ and $g$ in the integral, with maybe a few sign changes. Then try to approximate $f$ in $L^1$ by a smooth compactly supported function.

For instance, see

https://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/#more-4145

and

https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/

for more about the first suggestion.