My question is :
Let $f\in$$L^1(\mathbb{R})$, and define $g_n(x)$=$\frac{1}{n}\sum_{k=1}^{n}f(x+\frac{k}{n})$. Show that $g_n$ converges in $L^1(\mathbb{R})$ to some $F$ and find $F$.
First, I assume that $g_n$ converges to $F$ in $L^1(\mathbb{R})$ and $g$ is the pointwise limit of $g_n$. Then I can show that $F=g$ a.e. So, I try to find the pointwise limit of $g_n$. But I don't know how to find the pointwise limit of $g_n$.
For any $x\in\mathbb R$, consider the interval $[x,x+1]$. Then for a positive integer $n$, $$\left\{x, x+\frac1n, \ldots, x+\frac{n-1}n, x+1\right\} $$ is a partition of $[x,x+1]$, with each subinterval having width $\frac1n$. Therefore $g_n$ is actually a Riemann sum of $f$ over $[x,x+1]$. Since $f$ is Lebesgue integrable, it is Riemann integrable, and since $\frac1n\stackrel{n\to\infty}{\longrightarrow}0$, we have $$F(x)= \lim_{n\to\infty} g_n(x) = \int_x^{x+1} f(t)\mathsf dt. $$