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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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    With $f_n(x)=x^n$, we can see that only 3. or 4. can we true. Show that if $\{f_n\}$ are measurable, the maps $g_k(x):=\sup_{n\geq k}f_k(x)$ is measurable.2012-07-15

1 Answers 1

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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 $g_n^{-1}(-\infty,a]=\lbrace x :g_n(x)\leq a\rbrace=\lbrace x :\sup_{k\geq n}f_k (x)\leq a\rbrace=\lbrace x: f_k(x)\leq a\mbox{ for all } k\geq n\rbrace$ 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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    @LucM: You are right, thanks. Corrected.2015-03-12