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!

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    $\begingroup$ Prove it first for $f \in C([0,1])$ (it essentially follows from uniform continuity) and then estimate an $L^1$ function by a sequence of continuous functions. It is actually more difficult to prove convergence a.e. I think. $L^1$ convergence is much more natural here. $\endgroup$ – User8128 Oct 18 '16 at 4:29
  • $\begingroup$ Thanks. I haven't learnt how to estimate an $L^1$ function by continuous functions. Can it be done using Lusin theorem? $\endgroup$ – yoyostein Oct 18 '16 at 5:52
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    $\begingroup$ You can use Lebesgue's differentiation theorem to make your argument for pointwise a.e. convergence rigorous. $\endgroup$ – Dominique R.F. Oct 18 '16 at 6:09
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    $\begingroup$ I'm not sure Lusin's theorem will help here [though, of course, it might]. I'm simply talking about that fact that $C([0,1])$ is dense in $L^1([0,1])$. Thus for any $f \in L^1([0,1])$, we can find a sequence $f_n \in C^1([0,1])$ so that $\| f - f_n \|_{L^1([0,1])} \to 0.$ $\endgroup$ – User8128 Oct 19 '16 at 13:48

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.

  • $\begingroup$ Thanks a lot. When we are using Lusin Theorem, we also need this fact right: If $f$ is integrable, then for any measurable $|A|<\delta$, then $\int_A |f|<\epsilon$? How would we show that $P_nf$ is integrable? Essentially I can understand except the last line $\int_0^1 |P_nf-P_ng_\epsilon|<\epsilon$. Thanks a lot. $\endgroup$ – yoyostein Oct 25 '16 at 3:37
  • $\begingroup$ Just one more question: Why do we need $\int_0^1 |P_nf-P_ng_\epsilon|<\epsilon$? I don't see it used explicitly anywhere. $\endgroup$ – yoyostein Oct 25 '16 at 3:41
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    $\begingroup$ Think of $P_n f$ as $\sum a_k \chi_{k}$ and see what you when you integrate. $\endgroup$ – Jacky Chong Oct 25 '16 at 3:41
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    $\begingroup$ You need $\int^1_0 |P_n f-P_n g_\epsilon|<\epsilon$ because $\int |f-P_n f| \leq \int |f-g|+|g-P_ng|+|P_ng-P_n f|$. $\endgroup$ – Jacky Chong Oct 25 '16 at 3:42
  • $\begingroup$ Ok, very nice. Thanks and have a great week ahead! $\endgroup$ – yoyostein Oct 25 '16 at 3:49

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