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An exercise taken from Klenke: Let $f \in L^1(\lambda)$, where $\lambda$ is the restriction of the Lebesgue measure to $[0,1]$. Let $I_{n,k} = [k2^{-n},(k+1)2^{-n}), $ for $n \in \mathbb{N}$ and $k = 0,...,2^{n}-1$. Define $f_n:[0,1]\to \mathbb{R}$ by $$ f_n(x)=2^n\int_{I_{k,n}} f d\lambda$$ if k is chosen such that $x \in I_{k,n}$. Show that $\lim_{n\to \infty} f_n(x) = f(x)$ almost surely for $x \in [0,1]$.

But I'm not sure about here $$\lim_n f_n(x) = \lim_n 2^n\int_{k2^{-n}}^{(k+1)2^{-n}} f d\lambda...$$ since $\lim 2^n=0$? If we remove it then perhaps it makes sense below:

i.e. Say $\int_0^1f(x) dx=\int_{I_{k,n}}f \mathop{}\!d \lambda $, therefore $$ \lim_{n \to \infty}\int_0 ^1f(x) dx=\lim_{n \to \infty}\int_{I_{k,n}}f \mathop{}\!d \lambda \tag1 $$ Now as $\{I_{k,n}f\}$ is a family of non-negative functions, then the monotone convergence theorem yields

$$ \int_{[0,1]} f\mathop{}\!d \lambda =\lim_{n \to \infty}\int_{I_{k,n}}f \mathop{}\!d \lambda \tag2 $$

Then from $(1)$ and $(2)$ we have that $$ \int_{[0,1]} f\mathop{}\!d \lambda =\lim_{n \to \infty}\int_{k2^{-n}}^{(k+1)2^{-n}}f(x) \mathop{}\!d x\tag3 $$ So $f\in L^1$ if and only if $\lim_{n\to \infty}\int_{k2^{-n}}^{(k+1)2^{-n}}f(x) \mathop{}\!d x$ converges.

CCZ23
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    I am not sure about (2): looks like $\chi_{I_{k,n}} \not \to \chi_{[0,1]}$ as $n \to \infty$. Instead, this problem looks related to Lebesgue differentiation theorem. – mathdoge Jun 26 '20 at 17:33
  • If you want to investigate whether a formula is reasonable, you might start with a simple example, e.g. $f(x) = 1$. Using the given formula: $$2^n\int_{I_{k,n}} f d\lambda = 2^n\int_{I_{k,n}} 1 d\lambda = 2^n \lambda(I_{k,n}) = 2^n \cdot 2^{-n} = 1$$ Can you now see why the $2^n$ is appropriate, even necessary? – Brian Moehring Jun 26 '20 at 17:58

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First, $\lim_n2^n=\infty$. If you remove it, the limit of the integrals is zero. The point of the $2^n$ at the front is that it is $1/\lambda(I_{k,n})$.

Assume first that $f$ is continuous (so uniformly continuous). Given $\varepsilon>0$, there exists $\delta>0$ such that $|s-t|<\delta$ implies $|f(s)-f(t)|<\varepsilon$. So if $2^n>\tfrac1\delta$, then $|s-t|<\delta$ for all $s,t\in I_{k,n}$. Then \begin{align} |f(x)-f_n(x)| &=2^n\Bigg|\int_{I_{k,n}}(f(x)-f(t))\,d\lambda(t)\Bigg|\\[0.3cm] &\leq 2^n\,\int_{I_{k,n}}|f(x)-f(t)|\,d\lambda(t)\\[0.3cm] &\leq2^n\varepsilon\,\lambda(I_{k,n})=\varepsilon. \end{align} So $f_n(x)\to f(x)$ for all $x$.

Consider now a general $f\in L^1[0,1]$. Then there exists a sequence $\{g_n\}\subset L^1[0,1]$ with $\|g_n-f\|_1\to0$. From here we can choose a subsequence $\{g_{n_s}\}$ such that $g_{n_s} \to f$ almost uniformly. Now \begin{align} |f(x)-f_n(x)| &\leq |f(x)-g_{n_s}(x)|+\Big|g_{n_s}(x)-2^n\int_{I_{k,n}}g_{n_s}\Big|+\Big|2^n\int_{I_{k,n}}g_{n_s}-f_n(x)\Big|\\[0.3cm] &\leq|f(x)-g_{n_s}(x)|+\Big|2^n\int_{I_{k,n}}(g_{n_s}(x)-g_{n_s})\Big|+\Big|2^n\int_{I_{k,n}}(g_{n_s}-f)\Big|.\\[0.3cm] \end{align} The three terms go to zero: the first and third by the almost uniform convergence, and the middle one by continuity of $g_{n_s}$.

Martin Argerami
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