- #1

jinsing

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## Homework Statement

Suppose that [tex]f_n \stackrel{P}{\rightarrow} f[/tex] on [tex]S[/tex]. Given [tex]\epsilon >0[/tex] define [tex]E_{\epsilon,n}=\{x\in S \mid |f_n(x)-f(x)|<\epsilon\}.[/tex]

Prove that [tex]\forall \epsilon >0[/tex] [tex]\cup_{n=1}^\infty E_{\epsilon,n}=S[/tex].

## Homework Equations

Know about measurability, pointwise convergence etc.

## The Attempt at a Solution

I came up with this proof, but I'm not entirely sure it's right. It's also not very formalized..but besides that it almost seems like it's missing something:

Assume {f_n} -> f pointwise on S, and given epsilon>0 define E as above. Note that E = [tex] \bigcap_{n=1}^\infty \{x \mid |f_n(x) - f(x)| < \epsilon\}.[/tex]

Since each f_n is measurable and f_n -> f pointwise, then we know f_n-f is measurable for all n in N, and so [tex] \{x \mid f_n(x) - f(x)| < \epsilon\} [/tex] is measurable for all n in N, thus E_{n,\epsilon} is a countable intersection of measurable sets and therefore measurable.

Note that E_{n,epsilon} is contained in E_{n+1,epsilon} for all n in N. Note also that since f_n -> f pointwise for all x in S, f_n(x) -> f(x) as n->\infty. So for all x in S there exists a k in N such that whenever N\geq k |f_N(x) - f(x)| < \epsilon. Hence for all x in S, x is in E_{n,\epsilon} for some n in N. Thus [tex] \bigcup_{n=1}^\infty E_{n, \epsilon} = S,[/tex] as desired.

Thanks in advance, guys!

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