In class our professor wants us to give an example of a function $f$ on an interval $[a,b]$ and a sequence {$f_n$} converging to $f$ almost uniformly so that there is no set E of measure zero so that the sequence {$f_n$} converges uniformly to $f$ on $[a,b]\backslash E$.

I am a little confused about what he's asking; does this mean we want a function and sequence that does NOT converge uniformly almost everywhere to $f$? Or do we want a function in which uniform convergence (not just uniform convergence almost everywhere) holds?

  • $\begingroup$ Your first paragraph seems completely clear to me. $\endgroup$
    – zhw.
    Mar 16, 2016 at 1:03
  • $\begingroup$ If you want the symbol \ inside the dollar signs, you need \backslash. Or you can put \ outside the dollars. $\endgroup$ Mar 16, 2016 at 2:10

2 Answers 2


"almost uniformly" means that for any $\epsilon>0$, there exists a set $E$, $\mu(E)<\epsilon$ such that $f_n\to f$ uniformly on $E^c$. The name may be a little misleading. By Egorov's theorem, if $\mu(X)<\infty$, then a.s. convergence implies almost uniform convergence.

What you're asked to show that almost uniform convergence does not imply the stronger uniform convergence almost everywhere.

A simple example for this is the sequence $f_n=x^n$ on $[0,1]$. This converges pointwise to the function $0$ on $(0,1)$ and to $1$ at $0$, that is, it converges a.e. to the constant function $f=0$. The sequence clearly converges almost uniformly to $f$, because it converges uniforly to $0$ on every interval of the form $[0,1-\epsilon]$.

What about uniform convergence except on a set of measure zero ?

Let $E$ have zero measure. Observe that for any $m$, there exists a point in the interval $[1-1/m,1-1/(m+1)]$ not in $E$ (othwise $E$ would have positive measure). Therefore (using monotonicity of $f_nx^n$):

$$\sup_{x \in [0,1]\backslash E} |f_n (x) - f(x)|\ge (1-1/m)^n-0,$$

for any $m$. Letting $m\to \infty$, we see that

$$\sup_{x\in [0,1]\backslash E} |f_n (x)-f(x)|\ge 1-0=1.$$

(It is of course equal to $1$).

Thus, no matter which measure zero set $E$ we choose, we will never converge uniformly to the (a.e.) limit $f=0$ outside $E$.

  • $\begingroup$ I think you got a typo, it should be $f_n\rightarrow f$ uniformly on $E^c$. $\endgroup$
    – RJ Acuña
    Oct 13, 2021 at 7:11
  • 1
    $\begingroup$ Corrected. Thanks. $\endgroup$
    – Fnacool
    Oct 16, 2021 at 23:07

You might want to look up almost uniform convergence. It sounds like you want a function $f$ and a sequence of functions $\left\{f_n\right\}$ for which:

  1. Given any $\delta > 0$, there exists a set $A$, $m(A) < \delta$, such that $f_n \to f$ uniformly on $[a,b] \setminus A$
  2. Given any measure $0$ set $E$, the functions $f_n$ do not converge uniformly on $[a,b] \setminus E$.
  • 1
    $\begingroup$ Yes, I think that is what the professor asks $\endgroup$ Mar 16, 2016 at 1:18
  • $\begingroup$ Ah thanks! That makes a lot more sense. $\endgroup$
    – Min
    Mar 16, 2016 at 1:24

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