Lusin's Theorem - does $f$ ever have to be continuous? Recall Lusin's Theorem:

Let $f$ be a real-valued measurable function on $E=[0,1]$.  Then, for each $\epsilon > 0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\backslash F) < \epsilon$.

I was wondering if $f$ itself ever has to be continuous at any point (as a function on $E$)? I figured that it would at least need to be continuous on the interior of $F$, where it actually equals $g$ (I realize it technically equals $g$ on the boundary of $F$ as well, but we don't know what's going on outside $F$, so we can't necessarily say it's continuous on the boundary, can we?), because there it's coinciding with a function that is known to be continuous. 
Is there any more to this, and if so (and even if not), what would a proof look like?
 A: Since continuity (at a point) is a local property - whether $f$ is continuous at $x_0$ depends only on the values of $f$ on arbitrarily small neighbourhoods of $x_0$ - it indeed follows that $f$ must be continuous on the interior of $F$, if $f$ coincides with the continuous function $g$ on $F$.
But of course $F$ can have empty interior, so that doesn't imply that a measurable function needs to be continuous at any point.
And indeed, there are measurable functions that are nowhere continuous.
The characteristic function of $\mathbb{Q}\cap [0,1]$ is of course measurable and nowhere continuous.
However, by changing the values on a null set, we obtain the everywhere continuous function $0$. And conversely, every measurable function can be changed to a nowhere continuous measurable function by changing the values on a null set.
More interesting is the fact that there are measurable functions that are nowhere continuous and cannot be made continuous anywhere by changing the values on a null set.
The "simplest" example of such a function is the characteristic function of a measurable set $A$ with the property that $0 < m(A\cap I) < m(I)$ for all non-degenerate intervals $I$. Then every function $f$ that is a.e. equal to $\chi_A$ has points $x,y$ with $f(x) = 0$ and $f(y) = 1$ in every non-degenerate interval, hence cannot be continuous anywhere.
