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$f_n:[0,1]\to [0,1]$ be a continuous function and let $f:[0,1]\to [0,1]$ be defined by $$f(x)=\operatorname{lim\;sup}\limits_{n\rightarrow\infty}\; f_n(x)$$ Then $f$ is

  1. continuous and measurable

  2. continuous, but need not be measurable

  3. measurable, but need not be continuous

  4. need not be measurable or continuous.

I guess $3$ is correct, but I'm not able to prove it.

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This is a guess. What did you try to support it? –  Did Jul 15 '12 at 15:14
You mean Borel or Lebesgue measurable? –  Davide Giraudo Jul 15 '12 at 15:16
my intuition goes faster than my knowledge –  El Angel Exterminador Jul 15 '12 at 15:17
Borel Measurable –  El Angel Exterminador Jul 15 '12 at 15:18
Hint: The supremum of a set of measurable functions is measurable (as is the infinimum). Given this, can you express $f$ in terms of the inf of measurable functions? –  Benji Jul 15 '12 at 15:22

1 Answer 1

up vote 3 down vote accepted

Note that $$\lim\sup f_n(x)=\inf_{n\geq 1} \sup_{k\geq n} f_k(x)$$ Let $g_n(x)=\sup_{k\geq n}f_k(x)$, then $$g_n^{-1}(-\infty,a]=\lbrace x :g_n(x)\leq a\rbrace=\lbrace x :\sup_{k\geq n}f_k (x)\leq n\rbrace=\lbrace x: f_k(x)\leq a\mbox{ for all } k\geq n\rbrace$$ Hence $$g_n^{-1}(-\infty,a]=\bigcap_{k\geq n}f_k^{-1}(-\infty,a]$$ It follows that $g_n^{-1}(-\infty,a]$ is a measurable set (being intersection of measurable sets) and so $g_n$ is measurable.. In a similar fashion we can also prove that $\inf_{n\geq 1}g_n$ is also measurable (try), so we have $\lim\sup f_n(x)$ is measurable.

For the second part $f_n(x)=x^n$ converges pointwise to $g$ where $g(x)=0$ when $0\leq x<1$ and $g(1)=1$ and surely $g$ is not continous.

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Probably better to use $g_n$ rather than $f$ for the function, since it depends on $n$. –  Thomas Andrews Jul 15 '12 at 15:38
Yeah $g_n$ is better Thanks, I have changed it. –  pritam Jul 15 '12 at 15:44

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