Suppose that $f$ is continuous a.e. Let $\epsilon >0$. We know that for almost every $x$ there exists $\delta >0$ such that $|x<\mathrm{quote}/>-x|<\delta \Rightarrow |f(x<\mathrm{quote}/>)-f(x)|<\epsilon$. Let $A_n$ be the set of $x$ such that $\delta =1/n$ works. We see that $A_n$ is increasing (that is $A_n$ is contained in $A_{n+1}$ for all $n$) and the union of $A_n$ has measure 1.

Fix $0<c<1$. Since measure is a continuous set function we have ${\lim }_{n\to \infty }\mu (A_n)=\mu (A)=1$ (here $\mu (S)$ stands for the measure of the set $S$) so there exists $n$ such that $\mu (A_n)\ge c$.

In other words for any $\epsilon >0$ and any $0<c<1$ we can find an $n$ such that the set of $x$ such that:

$|x<\mathrm{quote}/>-x|<1/n\Rightarrow |f(x<\mathrm{quote}/>)-f(x)|<\epsilon$ (*)

has measure at least $c$.

Now take a dissection of [0,1] into intervals of length $<1/n$. The sum of the lengths of intervals which have an $x$ with the property (*) certainly is at least $c$ - since their union contains the set of $x$ with property (*) which has measure $c$. In each of these intervals there exists $x$ with the property (*) so $f$ varies by less than $2\epsilon$ in each of these intervals.

Now let's estimate the difference between the upper and lower sum. What about the contribution of the intervals which have an $x$ with (*)? Well $f$ varies by less than $2\epsilon$ in each of these intervals and the sum of the lengths of these intervals cannot exceed 1 so the maximum contribution to the difference between the lower and upper sum is $2\epsilon$. What about the contribution of the intervals which don't have an $x$ with property (*)? Well the sum of the lengths of these intervals is at most $(1-c)$ and $f$ varies by no more than $S-s$ in these intervals. (Where $S=\sup f$, $s=\inf f$.) Hence the maximum contribution is $(1-c)(S-s)$.

Hence the difference between the lower and upper sum is at most $2\epsilon +(1-c)(S-s)$.

To recap: given $\epsilon >0$ and $0<c<1$ we can find $n$ such that any dissection whose intervals have size $<1/n$ has the property that difference between the lower and upper sums is $<2\epsilon +(1-c)(S-s)$. This can be made arbitrarily small by taking $\epsilon$ small and $c$ close to 1. So we're done one way.

Suppose $f$ is not continuous a.e. Let $A_n$ be the set of $x$ such that: there doesn't exist $\delta >0$ with $|x<\mathrm{quote}/>-x|<\delta \Rightarrow |f(x)-f(x<\mathrm{quote}/>)|<1/n$. The union of the $A_n$ is the set on which $f$ is discontinuous. Since their union does not have measure zero there exists an $m$ such that $A_m$ does not have measure zero. In particular there exists a $c>0$ such that $A_m$ cannot be covered by disjoint intervals whose lengths sum to $<c$.

Dissect [0,1] into $n$ intervals of length $1/n$. $A_m$ must intersect at least $nc$ of these intervals (not counting intersections at endpoints). $f$ varies by at least $1/m$ on these intervals. So the difference between the upper and lower sums is $\ge 1/m\times nc\times 1/n=c/m$, for any $n$. If $f$ were Riemann integrable then the difference would tend to zero as $n\to \infty$ which it doesn't. Hence $f$ is not Riemann integrable.