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 '16 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 '16 at 2:10

"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 $E^c$. $\endgroup$ Oct 13 at 7:11
  • 1
    $\begingroup$ Corrected. Thanks. $\endgroup$
    – Fnacool

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 '16 at 1:18
  • $\begingroup$ Ah thanks! That makes a lot more sense. $\endgroup$
    – Min
    Mar 16 '16 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.