# Uniform and Pointwise convergence Almost Everywhere

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?

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

• I think you got a typo, it should be $f_n\rightarrow f$ uniformly on $E^c$. Commented Oct 13, 2021 at 7:11
• Corrected. Thanks. Commented 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 $[a,b] \setminus E$.
• Yes, I think that is what the professor asks Commented Mar 16, 2016 at 1:18
• Ah thanks! That makes a lot more sense.
– Min
Commented Mar 16, 2016 at 1:24