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Let $(\mathbb{N}, 2^{\mathbb{N}}, \mu)$ be a measure space, where $$\mu(E)= \sum_{n \in E} \frac{1}{2^n}$$ Prove that for each $\epsilon>0$ there exists $E \subset \mathbb{N}$ with $\mu(E^c)< \epsilon $, such that every pointwise convergent sequence $\{f_n\}$ converges uniformly on $E$.

I directly cannot use Egoroff's theorem directly as in the theorem $E$ depends on both $\epsilon$ and $\{f_n\}$, here I require $E$ to depend only on $\epsilon$.

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Suppose $f_n$ converges to $f$ pointwise. Let $E_k:=\{1,2,\cdots,k\}$. This means that for all $\delta>0$ and each $i\in E_k$, there is an $N_i$ such that $|f_n(i)-f(i)|<\delta$ for all $n\geq N_i$

Fix $\epsilon>0$. Pick any $E_k$ with such that $\mu(E_k^c)<\epsilon$ (e.g. take $k$ large enough). Then define $N:=\max_{i\in E_k}(N_i)$, which will give you uniform convergence on $E_k$.

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  • $\begingroup$ Can you please elaborate more on the last paragraph, are you using Egoroff? $\endgroup$ – Savannah Oct 3 '18 at 5:13
  • $\begingroup$ @Egoroff: No. I'm just using the fact that whatever $E_k$ you pick is finite, so you can always find an $N$ such that $|f_n(i)-f(i)|<\delta$ for all $n>N$ and all $i\in E_k$. $\endgroup$ – Alex R. Oct 3 '18 at 5:16
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Let $E=\{1,2,...,N\}$ where $N$ is so large that: $\sum_{n=N+1}^{\infty} \frac 1 {2^{n}} <\epsilon$. This set has the desired properties because on he finite set $E$ pointwise convergence implies uniform convergence.

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