$f(x+1/n)\to f$ in $L^1(\mathbb{R})$

Question

I'm working on a problem in my past Qual.

"Suppose $f:\mathbb{R}\to \mathbb{R}$ is in $L^1(\mathbb{R})$. Prove that $f(x+1/n)\to f$ in $L^1(\mathbb{R})$."

We have $\int |f(x+1/n)|=\int |f|<\infty$. The sequence is uniformly dominated by $f\in L^1$. So by the diagram in https://www.johndcook.com/blog/modes_of_convergence/, it suffices to prove that $f(x+1/n)\to f$ almost everywhere or in measure.

However, both involve evaluating $|f(x+1/n)- f(x)|$. I don't think this is possible since $f$ is not continuous. This is where I'm out of approaches.

Answer 1

Let $\epsilon >0$. Choose a continuous function $g$ with compact support such that $\int |f-g| <\epsilon /3$. Now $$\int |f(x+\frac 1 n)-f(x)| dx$$ $$\leq \int |f(x+\frac 1 n)-g(x+\frac 1 n)| dx$$ $$+\int |g(x+\frac 1 n)-g(x)| dx$$ $$+ \int|g(x)-f(x)| dx <2\epsilon /3+$$ $$\int |g(x+\frac 1 n)-g(x)| dx$$ To show that $\int |g(x+\frac 1 n)-g(x)| dx< \epsilon /3$ for $n$ sufficiently large use the following:

1) It is enough to integrate over a compact interval $[a,b]$

2) $g$ is uniformly continuous

Can you now finish the proof?