Convergence in $L^1$ of defined step functions Question: Let $f \in L^1(m)$. For k=1,2,... let f_k be the step function defined by
$$
f_k(x)=k\int_{j/k}^{(j+1)/k} f(t)\,dt
$$
$$
\text{for } \frac{j}{k}<x\leq\frac{j+1}{k}, \quad j=...,-1,0,1,...
$$
Show that $f_k$ converges to f in $L^1$.
I guess I showed $f_k$ converges f a.e. by the Lebesgue Differentiation Theorem, since
$$\lim f_k(x)=\lim \dfrac{\int_{j/k}^{(j+1)/k} f(t)\,dt}{1/k}$$
If I can also show $|f_k|\leq |g|$ a.e. for some $g\in L^1$ then I can use the Lebesgue Dominated Convergence Theorem since $|f_k-f|\leq |g|+|f|$ and $f_k-f$ converges to $0$ a.e. My attemt was to show $|f_k|\leq |f|$, but I'm not even sure whether this inequality holds or not.
Any comments on my idea and suggestions?
Thanks!
 A: You can prove this exactly as the standard result about approximate units in $L^1$.
Write $T_kf=f_k$. Then 
$$T_kf(x)=\int_{\mathbb R} \Phi_k(x,t)f(t)\, dt,$$
for a certain "kernel" $\Phi_k$ given by 
$$\Phi_k(x,t)=k\,\sum_{j\in\mathbb Z} \mathbf 1_{Q_{j,k}}(x,t)\, ,$$
where $Q_{j,k}$ is the square $(\frac{j}k,\frac{j+1}{k}]\times (\frac{j}k,\frac{j+1}{k}]\,$. 
The kernel $\Phi_k$ is nonnegative, and a direct computation shows that $\int_{\mathbb R} \Phi_k(x,t)dx=1$ for all $t$. Using Fubini's theorem just as for convolutions, this implies that the operators $T_k$ are bounded on $L^1$ with $\Vert T_k\Vert\leq 1$.
We also have $\int_{\mathbb R} \Phi_k(x,t)dt=1$ for all $x$, and it is not hard to check 
$$\lim_{k\to\infty} \int_{\{ t;\; \vert t-x\vert\geq\delta\}} \Phi_k(x,t)\, dt=0$$
for every fixed $\delta>0$, uniformly with respect to $x$. In fact, the above integral is equal to $0$ for all $x$ whenever $k>1/\delta$, since in this case $(x,t)$ does not belong to any $Q_{j,k}, j\in\mathbb Z$ if $\vert x-t\vert\geq \delta$. Using this, one shows in the "usual" way that $T_kf\to f$ in $L^1$ when $f$ is continuous with compact support. So the result holds for all $f\in L^1$ since the operators $T_k$ are uniformly bounded.
