We say a function $f: \mathbb R^n \to \mathbb R$ is essentially continuous if for every $x \in \mathbb R^n$, we have
$$\lim_{\delta \to 0_+} \text{esssup}_{y, z \in B_\delta (x)} |f(y) - f(z)| = 0,$$
where the essential supremum is taken with respect to the product Lebesgue measure.
Question: Let $f$ be essentially continuous. Does there necessarily exist some classically continuous function $f^{\ast}$ that agrees Lebesgue almost everywhere with $f$?
I mean sure. First, for each $x$ the essential limit of $f(y), y\to x$ exists, because for each $\delta$ we get an interval of potential values (between essinf and esssup), these intervals shrink by assumption, hence the limit exists as the intersection point of these intervals. Call this function $f^*(x)$. We need to show that this function is continuous and $f(x) = f^*(x)$ a.e.
For the first one, for arbitrary $x$ and $\varepsilon > 0$ the interval for $x$ contains the intervals for all points at most $\varepsilon$ away from $x$, so the limiting value for them as well, and since these intervals shrink when $\varepsilon \to 0$, we get continuity.
For the second one, if not then there is $\delta > 0$ such that $|f(x)-f^*(x)| > \delta$ on a compact set of positive measure. This set has a Lebesgue point $x$, say, in the vicinity of which we have a bigger and bigger intersection, in particular it is always of positive measure. But this contradicts already proven facts that $f$ has essential limit $f^*(x)$ at $x$ and $f^*$ is continuous there, so $f^*(y)$ tends to the same limit.
Denote by $f_{\delta}(x) = \frac{1}{|B_{\delta}(x)|} \int_{B_{\delta}(x)} f$ the scale $\delta$ mean of $f$ centered at $x$.
Let $\epsilon > 0$, then there exists $\delta > 0$ such that $$ \operatorname{esssup}_{y,z\in B_{\delta}(x)} |f(y) - f(z)| < \epsilon \tag{E}$$ Given $\delta_1, \delta_2 < \delta$, we can take averages over $y\in B_{\delta_1}(x)$ and $z\in B_{\delta_2}(x)$, to find $$ | f_{\delta_1}(x) - f_{\delta_2}(x) | < \epsilon $$ This shows that as $\delta\to 0$ the net $f_\delta(x)$ is Cauchy, and hence converges. Let
$$ g(x) = \lim_{\delta\to 0} f_{\delta}(x) $$
I claim that $g$ is continuous.
Let $\epsilon > 0$, and $\delta > 0$ such that (E) holds. For $y \in B_{\delta}(x)$, choose $\delta'$ such that $B_{\delta'}(y)\subseteq B_{\delta}(x)$ and $|f_{\delta'}(y) - g(y)| < \epsilon$. Choose $\delta'' < \delta$ such that $|f_{\delta''}(x) - g(x)| < \epsilon$. Then
$$ |g(x) - g(y)| \leq |f_{\delta'}(y) - f_{\delta''}(x)| + |g(y) - f_{\delta'}(y)| + |g(x) - f_{\delta''}(x)| < 3 \epsilon$$
The latter two terms are $ < \epsilon$ by convergence of $f_{\delta}$ to $g$. The first term is $<\epsilon$ because it is obtained by averaging (E), and that $B_{\delta'}(y)$ and $B_{\delta''}(x)$ are both contained in $B_{\delta}(x)$ by assumption.
Finally, we have that $g = f$ a.e. by Lebesgue differentiation.
