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Consider a sequence $(f_n)$ defined on $\mathbb{R}$ by $f_n =\chi_{[n,n+1]},$ $n\in \mathbb{N}$ and the function $f\equiv 0.$ Does $f_n$ converge to $f$ almost everywhere, almost uniformly or in measure?

$f_n\to f$ almost everywhere is the same as saying that $f_n\to f$ pointwise almost everywhere, i.e. on a subset whose complement has measure zero. Given $\epsilon>0$ and $x\in \mathbb{R}$ we observe that if $x\geq 0$ then for some $n_0\in \mathbb{N}$ we have that $n_0\leq x<n_0+1.$ And so if we choose $N=n_0+1$ then for all $n\geq N$ we have that $f_n(x)=0$ and so $$|f_n(x)-0|<\epsilon.$$ On the other hand if $x<0$ then $f_{n}(x)=0$ and so we see that $$\lim_{n\to \infty}f_n(x) = f(x)$$ for all $x\in \mathbb{R}.$ Since the complement of $\mathbb{R}$ is $\emptyset$ which has measure $0$ we conclude that $f_n$ converges to $f$ almost everywhere.

Now $f_n$ does not converge uniformly to $f$ on the set $[0,\infty),$ a set of infinite measure. And so we conclude that $f_n$ does not converge almost uniformly to $f.$

If we choose $\epsilon = 0$ and $\eta =5$ then for all $N(\epsilon,\eta)\in \mathbb{N}$ we have that for $n\geq N.$ $$\mu(x\in D:|f_n|\geq \epsilon)=\mu(\mathbb{R})=+\infty\geq 5.$$

So we see that $f_n$ does not converge to $f$ in measure.

Is this solution correct?

This is problem 39 page 48 on the following pdf https://huynhcam.files.wordpress.com/2013/07/anhquangle-measure-and-integration-full-www-mathvn-com.pdf

The definition for convergence in measure used there is:

enter image description here

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  • $\begingroup$ what is $\eta$? the measure of the set where $f_n$ disagrees with $f$ only has lebesgue measure 1. Not sure where the $\mu(|f_n| > \varepsilon) = +\infty$ comes from. $\endgroup$ – Daniel Xiang Sep 5 '18 at 2:51
  • $\begingroup$ Hi, I made some changes. Please refer to the new post. $\endgroup$ – model_checker Sep 5 '18 at 3:01
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You are correct that $f_n$ does not converge to $f$ in measure, though your reasoning is flawed. In the definition you provide, it says "for any $\varepsilon > 0, \dots$". You've chosen $\varepsilon = 0$. If you instead pick $\varepsilon = \eta = 0.5$, then \begin{align*} \mu(|f_n-f| > \varepsilon) = \mu([n,n+1)) = 1 > 0.5 = \eta. \end{align*} So no matter how large $N$ is, there will always exist an $n > N$ for which $\mu(|f_n-f| > \varepsilon) > \eta$.

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