# Converge pointwise a.e. but not converge uniformly a.e.?

Show that on space $$([0,1],\lambda,\mathcal{L})$$ where $$\lambda$$ is Lebesgue measure defined on Lebesgue $$\sigma-$$ algebra $$\mathcal{L}$$ .

function $$f_n(x) =x^n$$ on $$[0,1]$$ is not uniformly converge to $$f =0$$ almost everywhere.

It's easy to see that $$x^n$$ converge uniformly to $$0$$ on $$[0,a]$$ for any fixed $$a<1$$. But I can't figure out why there is not any zero measure set $$E^c$$such that on its complement $$f_n(x)$$ converge uniformly to $$0$$.

• Do you mean the sequence $f_n(x)=x^n$? Commented May 9, 2020 at 6:16
• @Robert Shore , thanks, I forgot to type it. Commented May 9, 2020 at 6:17
• As $E$ has full measure you find for any $\varepsilon>0$ that $(1-\varepsilon,1)\cap E\neq \emptyset$. This tells you with the same argument as you linked, that $f_n\mid_E$ does not converge uniformly to zero. Commented May 9, 2020 at 6:25

Suppose $$E^{c}$$ has measure $$0$$ and $$x^{n} \to 0$$ uniformly on $$E$$. Then $$E$$ is dense. Hence there exists a point $$x_n \in (1-\frac 1 n, 1) \cap E$$. Now $$sup \{x^{n}: x \in E\} \geq x_n^{n} >(1-\frac 1n)^{n} \to \frac 1 e$$, a contradiction.