# Variant of Egoroff's theorem

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 $$\epsilon$$.

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 $$E_k$$.
• @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 $i\in E_k$. – Alex R. Oct 3 '18 at 5:16
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.