Let $f ∈ L_1(\mathbb{R}).$ For $n ∈ \mathbb{N}$ define the function $g_n :\mathbb{R}→\mathbb{R}$ as follows. For $k ∈ \mathbb{Z}$ and for $x ∈ [k/n, (k + 1)/n)$ set $g_n(x) = n\int_{k/n}^{\frac{k + 1}{n}}f(x)dx$. Prove that $g_n$ converges to $f$ a.e. and in $L_1(R).$
I proved the converges a.e. using Lebesgue Differentiation Theorem. Not sure how to do convergence in $L_1$.
You can proceed exactly as in the link given by PhoemueX. Here are some details.
First, observe that $\Vert g_n\Vert_1\leq \Vert f\Vert_1$ (this is quite easy to check). This means that if you set $T_n(f)=g_n$, then $T_n$ is a bounded (linear) operator from $L^1$ into itself, with $\Vert T_n\Vert\leq 1$.
Since the $T_n$ are uniformly bounded, it is enough to show that $T_n(f)\to f$ for avery $f$ in some dense subspace $\mathcal E$ of $L^1$.
Now, you have a very good candidate: consider the linear span $\mathcal E$ of all functions $\mathbf 1_{I_{k,n}}$, $n\in\mathbb N$, $k\in\mathbb Z$, where $I_{k,n}$ is the interval $[\frac{k}n,\frac{k+1}n)$.
It is rather clear that $\mathcal E$ is dense in $L^1$. So you just have to check (by linearity) that for any fixed $n_0, k_0$, the result holds for $f=\mathbf 1_{I_{k_0,n_0}}$. This should cause you no trouble.