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

$$
f_k(x)=k\int_{j/k}^{(j+1)/k} f(t)\,dt
$$
$$
\text{for } \frac{j}{k}&lt;x\leq\frac{j+1}{k}, \quad j=...,-1,0,1,...
$$

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!
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 $\delta&gt;0$, uniformly with respect to $x$. In fact, the above integral is equal to $0$ for all $x$ whenever $k&gt;1/\delta$, 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.
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 $\delta&gt;0$, uniformly with respect to $x$. In fact, the above integral is equal to $0$ for all $x$ whenever $k&gt;1/\delta$, 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.