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Need a little help with this {f_n} converging to f(x) proof

  • Thread starter jinsing
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  • #1
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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!
 
Last edited:

Answers and Replies

  • #2
382
4
Imo this is a bad question. There is nothing to prove just write down the definition of pointwise convergence.

let x be an element of x. By Assumption for all ε>0 there exists M s.t. if n>M |fn(x) - f(x)| <ε. Hence x is an element of En for some n.
 
  • #3
30
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Yeah, it seemed suspiciously straight-forward to me too. I guess I was wondering if there was any discrepancy between the 'given epsilon > 0' and having to prove 'for all epsilon > 0.' Is this the same thing?
 

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