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 $0$.

A related question:A sequence of continuous functions on [0,1] which converge pointwise a.e. but does not converge uniformly on any interval

  • $\begingroup$ Do you mean the sequence $f_n(x)=x^n$? $\endgroup$ Commented May 9, 2020 at 6:16
  • $\begingroup$ @Robert Shore , thanks, I forgot to type it. $\endgroup$
    – yi li
    Commented May 9, 2020 at 6:17
  • 1
    $\begingroup$ 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. $\endgroup$ Commented May 9, 2020 at 6:25

1 Answer 1


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.


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