About the converge of simple functions I am now considering such a problem.
Suppose $f\in L^1([0,1])$, for any integer $n$, we can construct such a measurable function:
$$Pn(f)=n\sum\limits_{k=1}^n\int_\frac{k-1}{n}^\frac{k}{n} f d\lambda\cdot \chi_{[\frac{k-1}{n},\frac{k}{n}]}$$
By computing, I have found for any $n\in\mathbb{N}$, $P_n(f)$ is Lebesgue measurable, and the integral of which is equal to $\int_{[0,1]} fd\lambda$. My question is how to prove $P_n(f)$ is converge to $f$ in $L^1$, i.e.
$$\lim\limits_{n\rightarrow\infty}\int_{[0,1]}|P_n(f)-f|d\lambda=0$$
I tried to use LDCT to show that, however, I can not find a correct dominating function, and I have difficulty in proving $Pn(f)$ converges to $f$ pointwise. Wish you could give me some hints.  
 A: As you suggest in a comment, you can prove this by an $L^1$ approximation argument. It is obviously true, for example, for Riemann integrable functions.
A very similar exercise is to show $\tau_x f\to f$ in $L^1(\mathbb R)$ as $x\to 0,$ where $\tau_x$ is the translation operator $\tau_x(f)(y)=f(y+x),$ and you can use an almost identical argument as in this answer: https://math.stackexchange.com/q/458234
Dominated convergence cannot be used directly. If we take $f=1/x\log(x)^2$ then $f\in L^1([0,1]),$ but $\sup_nP_n(f)(x)$ is roughly $1/x\log(x)$ which is not integrable on $[0,1]$. In other words the Hardy maximal function may not lie in $L^1([0,1]).$

Just an aside. If we restrict $n$ to powers of two, the result follows immediately from the $L^1$ Martingale Convergence Theorem. The full result follows immediately from a Martingale Convergence Theorem for derivations, because the partitions of $[0,1]$ have the "Vitali property" as discussed for example in Stopping Times and Directed Processes by G. A. Edgar and Louis Sucheston.
