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? 
 A: 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:


*

*Given any $\delta > 0$, there exists a set $A$, $m(A) < \delta$, such that $f_n \to f$ uniformly on $[a,b] \setminus A$

*Given any measure $0$ set $E$, the functions $f_n$ do not converge uniformly on $[a,b] \setminus E$.

A: "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 $E$.
