Is the set $\{x\in \mathbb{R}: \lim_{y\rightarrow x}f(y)$ exist$\}$ measurable? 
Let $f$ be a Lebesgue Measurable Function on $\mathbb{R}$. 
  Is the set $\{x\in \mathbb{R}: \lim_{y\rightarrow x}f(y)$ exist$\}$ always measurable?

I can prove if $a\in \mathbb{R}$ is given $\{x\in \mathbb{R}: \lim_{y\rightarrow x}f(y)=a\}$ is measurable by showing the $\epsilon-\delta$ definition of limit only requires $\frac{1}{n}$ values($n\in \mathbb{N}$) for $\epsilon$ and $\delta$ and write the set as countable union/intersection of measurable sets. However I don't think we can do the same thing for $a$ becuase it has uncountably many possibilities.
 A: We will consider a variant of the oscillation; first, for $x\in\mathbb R$ and $\delta>0$, define $I(x,\delta)=(x-\delta,x)\cup(x,x+\delta)$, and let $$\omega_{\delta}(x)=\sup_{I(x,\delta)}f-\inf_{I(x,\delta)}f.$$ Define also $$\omega(x)=\limsup_{n\to\infty}\omega_{1/n}(x).$$ Suppose that $\lim_{y\to x}f(y)=a\in\mathbb R$ for some $x$, and let $\varepsilon>0$. Then, there exists $N\in\mathbb N$ such that, if $y\in I(x,1/m)$, for $m\geq N$, then $|f(y)-a|<\varepsilon$. This shows that $\omega_{1/m}(x)\leq 2\varepsilon$ for all $m\geq N$, therefore $\omega(x)=0$.
Conversely, if $\omega(x)=0$, we can show that $f$ has a limit at $x$. Therefore, the set of points at which $f$ has a limit is equal to the set of roots of $f$.
It will be enough now to show that $\omega$ is measurable. From its definition, it is enough to show that $\omega_{1/n}$ is measurable for $n\in\mathbb N$, therefore it is enough to show that $$\sup_{I(x,1/n)}f,\,\,\inf_{I(x,1/n)}f$$ are measurable, as functions of $x$, which is true. Therefore, the set of points for which the limit of $f$ exists is measureable.
