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Thread: uniform convergence

  1. #1
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    uniform convergence

    Let $\displaystyle f_n= n \chi_{[\frac{1}{n},\frac{2}{n}]} \,$ on $\displaystyle \mathbb{R}$ for all $\displaystyle n \in \mathbb{N}$ and $\displaystyle f=0 $ on $\displaystyle \mathbb{R}.$
    Show that ther does not exist a set of measure zero ,on the complement of which $\displaystyle (f_n)$ is uniformly convergent.
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  2. #2
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by problem View Post
    Let $\displaystyle f_n= n \chi_{[\frac{1}{n},\frac{2}{n}]} \,$ on $\displaystyle \mathbb{R}$ for all $\displaystyle n \in \mathbb{N}$ and $\displaystyle f=0 $ on $\displaystyle \mathbb{R}.$
    Show that ther does not exist a set of measure zero ,on the complement of which $\displaystyle (f_n)$ is uniformly convergent.
    Your functions $\displaystyle f_n$ are converging pointwise to the zero function. If you had some set $\displaystyle E$ as above, then you'd need, for any $\displaystyle \varepsilon>0$, that there exists some $\displaystyle N_\varepsilon$ so that $\displaystyle |f_n|<\varepsilon$ on $\displaystyle E$ for all $\displaystyle n>N_\varepsilon$. But if $\displaystyle E$ has measure zero, then the complement contains points arbitrarily near zero, i.e. points $\displaystyle x$ such that $\displaystyle x\in[\frac1n,\frac2n]$ for $\displaystyle n$ as large as you like, which contradicts uniform convergence.
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