Limit of Integral of "Narrowing Limits" Function Let $f$ be Lebesgue integrable on $[0,1]$, and define $$\large P_nf=n\sum_{k=1}^n\int_{(k-1)/n}^{k/n}f\,d\mu\cdot\chi_{[\frac{k-1}{n},\frac{k}{n}]}.$$
How do we show that $$\lim_{n\to\infty}\int_0^1 |f-P_nf|\,d\mu=0$$?
Thank you very much.

My attempts:
I would think in the direction of either Dominated Convergence Theorem, Monotone Convergence Theorem, or Fatou's Lemma (these are the only 3 tricks I know).
Firstly, I would like to prove $P_nf\to f$ a.e. I can see it intuitively, but not sure how to prove it. I can see that $P_nf$ is trying to partition $[0,1]$ into $n$ subintervals, then for any $x$, it is in one of the intervals $[\frac{k-1}{n},\frac{k}{n}]$, so that $P_nf(x)=n\int_{(k-1)/n}^{k/n}f(x)\,d\mu\to n\cdot\frac{1}{n}f(x)=f(x)$. (How to show this rigorously?)
I can't seem to use DCT due to the "$n$" in $P_nf$ which grows to infinity, so it is hard to find an explicit dominating function. Similarly, MCT seems hard since it is hard to show that $|P_nf|$ is increasing ($n$ is increasing but the "narrowing limits integral" may be decreasing) . That leaves Fatou's Lemma, and I am stuck.
Thanks for any help once again!
 A: A straight forward computation yields
\begin{align}
\int^1_0|f-P_n f|\ d\mu =&\ \int^1_0\left|\sum^n_{k=1} f(x)\chi_{[\frac{k-1}{n},\frac{k}{n}]}-n\sum^n_{k=1} \int^{k/n}_{(k-1)/n} f\ d\mu\cdot \chi_{[\frac{k-1}{n},\frac{k}{n}]}\right|\ d\mu\\
\leq&\ \sum^n_{k=1}\int^{k/n}_{(k-1)/n}\left|f(x)-n\int^{k/n}_{(k-1)/n} f\ d\mu \right|\ d\mu\\
=&\ n\sum^n_{k=1}\int^{k/n}_{(k-1)/n}\left|\int^{k/n}_{(k-1)/n} f(x)-f(y)\ d\mu(y) \right|\ d\mu(x)\\
\leq&\ n\sum^n_{k=1} \int^{k/n}_{(k-1)/n}\int^{k/n}_{(k-1)/n}|f(x)-f(y)|\ d\mu(x)d\mu(y). 
\end{align}
If $f$ is continuous on $[0, 1]$, then it's uniformly continuous on $[0, 1]$, which mean for each $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$.
In our case since $x, y \in [(k-1)/n, k/n]$, then for big enough $N$ we have that $|x-y|<\delta$. Hence it follows
\begin{align}
 n\sum^n_{k=1} \int^{k/n}_{(k-1)/n}\int^{k/n}_{(k-1)/n}|f(x)-f(y)|\ d\mu(x)d\mu(y) < \epsilon n\sum^n_{k=1} \int^{k/n}_{(k-1)/n}\int^{k/n}_{(k-1)/n} d\mu(y)d\mu(x) = \epsilon.
\end{align}
Thus, if we could approximate $f$ by a continuous function then we are essentially done. 
Fix $\epsilon>0$. By Lusin's Theorem, there exists $g_\epsilon$ continuous such that
\begin{align}
\int^1_0 |f-g_\epsilon|<\epsilon
\end{align}
and likewise we have
\begin{align}
\int^1_0 |P_n f-P_n g_\epsilon|<\epsilon. 
\end{align}
Important Note: the $g_\epsilon$ that you get from Lusin's theorem is actually equal to $f$ expect on a set of small measure. 
