Suppose I have a function $f_n$ = $\frac1n$ $\chi_{[-n,n]}$. I read somewhere that this function is certainly uniformly convergent but not convergent in measure. How is that? Can I get a better explanation? I would really appreciate some input on this.
Relationship between uniform convergence and convergence in measure
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0Think about why dominated convergence fails in your example. – 2011-03-28
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2You mean a sequence of functions, not a function. Do you know the definitions of uniformly convergent and convergent in measure? – 2011-03-28
1 Answers
First of all, you don't have one single function, but a sequence of functions, namely $f_{n} = \frac{1}{n} \chi_{[-n,n]}$. This sequence converges uniformly to the function $f = 0$ because for all $\varepsilon \gt 0$ there exists $n_{0}$ such that $\frac{1}{n_{0}} \lt \varepsilon$, hence $\Vert f_{n} - f\Vert_{\infty} = \operatorname{ess\,sup}_{x \in \mathbb{R}} |f_{n}(x) - f(x)| = \frac{1}{n} \leq \frac{1}{n_{0}} \lt \varepsilon$ for all $n \geq n_{0}$.
Second, it is true that any sequence of functions $g_{n}$ converging uniformly to $g$ also converges to $g$ in measure. Recall what convergence in measure means: For all $\varepsilon \gt 0$ we have $\lim_{n \to \infty} \mu(\{x \in \mathbb{R}\,:\,|g_n(x) - g(x)| \geq \varepsilon\}) = 0$. The implication holds because by uniform convergence we have $\Vert g_{n} - g\Vert_\infty \lt \varepsilon$ for $n$ large enough, so the measure of the set in question is actually zero for such $n$. It is therefore impossible to converge uniformly but not in measure.
However, the sequence $f_{n}$ does not converge in $L^1$. First of all, if there existed $f \in L^1$ with $f_{n} \to f$ in $L^1$ then there would be a subsequence $f_{n_j}$ converging pointwise a.e. to $f$, so necessarily $f = 0$. On the other hand, $\Vert f_n - f\Vert_1 = \int |f_{n} - f| = 2 \gt 0$, a contradiction. I leave it to you to check that for $p \gt 1$ the sequence $f_{n}$ converges to $f = 0$ in $L^p$ .
Further discussion of these and more topics can be found on John D. Cook's neat summary here.
In accordance with cardinal's new title to the question, I should point out the example $h_n = \chi_{[0,\frac{1}{n}]}$ which converges to $0$ pointwise and in measure but clearly not uniformly.
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3Another proof, perhaps more direct (without using any subsequence), that $(f_n)$ does not converge in $L^1$ is the following. For all $n,p \in \mathbb{N}$ one has $\lVert f_n-f_{n+p}\rVert_1=\frac{4p}{n+p}$ and the RHside is s.t. $\lim_{p\to \infty} \frac{4p}{n+p}=\sup_{p\in \mathbb{N}} \frac{4p}{n+p}=4$; thus $(f_n)$ is not Cauchy's in $L^1$, therefore $(f_n)$ does not converge. – 2011-03-28
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1On a related note, this function also makes a good example of the failure of the bounded convergence theorem when the measure of the underlying space is infinite. – 2011-03-28
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0@Pacciu: Yes, you're right that it is more direct. Even more direct would be to consider $\Vert f_{n} - f_{2n}\Vert_1 = \frac{4}{3}$ (provided your calculation is correct). On the other hand, I find it easier in practice to observe that there is already a pointwise limit, so only one candidate for an $L^1$-limit. – 2011-03-28
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0@cardinal: thanks, I'm getting tired (as is also shown in my silly mistake in my comment to Pacciu, where I should have written $\frac{4}{2}$). – 2011-03-28
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0@Theo: no worries. I deleted my response as well, since it is out of place without the context. – 2011-03-28
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0@Theo: I think it is correct, for the area under $|f_n-f_{n+p}|$ is nothing but a union of rectangles... Hence it is really really hard to miscalculate $\lVert f_n-f_{n+p}\rVert_1$! XD And for $p=n$ one has $\lVert f_n -f_{2n}\rVert_1= \frac{4n}{n+n}=2$. :-P – 2011-03-28
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0@Pacciu: :D Yeah, rub it in! :s Good night y'all. – 2011-03-28