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Let $f$ be a real valued function then the set where $f$ is continuous is measurable.

Note the statement does not even requires that $f$ is measurable.

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Hint: $f$ is continuous at $x$ if and only if $x\in \cap_{n=1}^\infty O_n$, where $O_n=\{x | \text{ there is a }\delta_x>0\text{ such that } |f(y)-f(z)|<n^{-1}, \text{ whenever }y,z\in(x-\delta_x,x+\delta_x) \}$. –  David Mitra Feb 24 '12 at 2:14

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Recall that the oscillation of $f$ at $x$ is defined by $osc_f(x)=\inf\{diam(f(-\frac 1 n+x,x+\frac 1 n)): n\in\mathbb N\}$. Let $O_n=\{x: osc_f(x)<\frac 1 n\}$. Note that $f$ is continuous at $x$ iff $osc_f(x)=0$ and $O_n$ is open (thus measurable) for all $n\in O_n$. Therefore $\bigcap_n O_n$ is the set of points where $f$ is continuous and it is measurable since is a countable intersection of measurable sets.

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