Relationship between uniform convergence and convergence in measure 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.
 A: 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.
