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Let $f$ be the indicator function of the interval $[0,1]$.

$f$ is clearly integrable, but how does one prove that there is no continuous function $g$ such that $g = f$ almost everywhere?

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    Just out of interest, why would you want to prove that? – 2017-01-20
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    @man_in_green_shirt I am self-studying measure theory and as I was reading the notion of almost everywhere, that came to my mind – 2017-01-20

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Let $g$ be a function which equals $f$ almost everywhere. This means that for any $n\in\mathbb N$ there exists $x_n\in(-1/n,0)$ for which $g(x_n)=f(x_n)=0$ (otherwise for some interval $(-1/n,0)$ functions $f$ and $g$ would be distinct there, and it's measure is positive). The same is for some $y_n\in(0,1/n)$, so $g(y_n)=f(y_n)=1$. Then we have \begin{align*} &\lim_{n\to\infty}x_n=0,\quad \lim g(x_n)=0,\\ &\lim_{n\to\infty}y_n=0,\quad \lim g(y_n)=1, \end{align*} hence, $g$ is discontinuous in zero.