Here is a tentative solution. Thank you Robert.
Problem Statement: A function is measurable if and only if it is the a.e. limit of continuous functions.
Proof: The reverse direction is trivial (see original post).
To prove the converse, we begin with proving a sequence of weaker statements which lead to the conclusion (since we want to apply certain theorems which require stronger hypotheses on $f$, and also because it significantly simplifies the notational mess that would occur if we had to deal with all exceptional cases at once). First, suppose $f$ is measurable, $f$ is defined on a set $A\subset\mathbb{R}^{d}$ of finite measure, and $f$ is finite on $A$. Then Lusin's Theorem guarantees us the existence of a compact $B\subset A$ where $m(A-B)\leq\epsilon$ for all $\epsilon>0$ and $f_{B}$ is uniformly continuous ($f_{B}$ is the "restriction" of $f$ to $B$; it is not defined on $A$!). An application of the Tietze Extension Theorem allows us to construct a new function $F$ such that $F$ is continuous on all of $A$ and $F=f_{B}$ on $B\subset A$ ($F$ is the continuous extension of $f_{B}$ from the subset $B$ to $A$). There are now many ways to construct a sequence of continuous functions. One trivial example is $\{f_{n}\}_{n=1}^{\infty}$ where $f_{n}=F\;\forall\;n=1,2,3\ldots$. A less trivial one comes from an application of the Stone-Weierstrass Approximation Theorem which guarantees the existence of a sequence of continuous polynomials which converge uniformly to $F$ on any compact set (take the closure of $A$ if necessary). In any case, we have a sequence of continuous functions $\{f_{n}\}_{n=1}^{\infty}$ where $f_{n}\to F$ on $A$. And since $F=f_{B}$ on $B$ is equivalent to $F=f$ on $B$, we have that $f_{n}\to f\;a.e.\;x\in A$ since $m(A-B)\leq\epsilon$.
Now weaken the hypotheses slightly by allowing $f$ to be infinite on a set of measure zero. Then we can let $B\subset A$ be the set on which $f$ is finite, where $m(A-B)=0$. Then from the above proof, we see there is a sequence of continuous functions $\{f_{n}\}_{n=1}^{\infty}$ which converges to $f\;a.e.$ on $B$. Let $C\subset B$ be the subset on which the convergence occurs. Then $f_{n}\to f$ on $(A-D)$ where $D=(B-C)\cup(A-B)$ is the union of two sets of measure zero. Hence, $f_{n}\to f\;a.e.\;x\in A$ since $D$ being the union of two sets of measure zero is again a set of measure zero.
Weakening the hypotheses further, we now allow the possibility of $E$ having non-finite measure, in particular, $E$ can be unbounded. Then $A=\bigcup_{n=1}^{\infty}A_{n}$ where $A_{n}=A\cap B_{n}$ and $B_{n}$ is a ball of radius $n$ about the origin. Since $\{B_{n}\}_{n=1}^{\infty}\nearrow\mathbb{R}^{d}$, for any subset of $A$ you desire $a.e.$ convergence, there exists an $N$ such that an application of the above proofs yield the desired result on $A_{N}$, and letting $N\to\infty$ gives the required conclusion on all of $A$.
Finally, we weaken the hypotheses to the original statement. In particular, we now allow $f$ to be infinite on a set of positive measure. To prove the same result holds, consider "cutoff" functions $g_{n} = f$ on the set where $f$ is finite, and $g_{n}=n$ on the set where $f$ is infinite. Then each $g_{n}$ is measurable, and so the above proofs apply to $g_{n}$ for all $n=1,2,\ldots$ Evidently $g_{n}\to f$ as $n\to\infty$, and so the sequences of continuous functions converging $a.e.$ to each $g_{n}$, also converge to $f$ at points where $f$ is finite, and as $n\to\infty$, to arbitrarily large values at points where $f$ is infinite. We therefore conclude the theorem holds when $f$ is infinite on sets of positive measure, thus completing the proof. (Note, in the final two cases, the index $n$ has nothing to do with the index on sequences of functions; it in fact indexes a collection of sequences of functions). QED
NOTE: I'm a little unsure about the final two cases; I was trying to apply Robert's hints, but I think I'm not being technical enough to offer a rigorous proof. Modifications are welcomed.
For brevity, write
$$\Delta(x,h) := \frac{J(x+h)-J(x)}{h} \qquad \text{and} \qquad \Delta_n(x,h) := \frac{J_n(x+h)-J_n(x)}{h} $$
Step 1: By definition of limsup, we have
$$\begin{align*} \limsup_{h \to 0} \Delta(x,h) > \epsilon &\iff \forall m \in \mathbb{N} \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \{0\}: \Delta(x,h) > \epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right): \Delta(x,h)>\epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m: \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon. \end{align*}$$
This shows that
$$\{x; \limsup_{h \to 0} \Delta(x,h)>\epsilon\} = \bigcap_{m \in \mathbb{N}} \bigcup_{k \geq m} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\}.$$
Consequently, it suffices to prove that for each fixed $k,m \in \mathbb{N}$ and $\epsilon>0$ the set
$$A := A_{k,m,\epsilon} := \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\} \tag{1}$$
is measurable.
Step 2: We claim that $$A = \bigcup_{\ell \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell} \bigg\}. \tag{2}$$
For brevity of notation, we set $$I := I_{k,m} := \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right).$$
Proof: Let $x \in A$, then $\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon$. By the definition of $\sup$, we can find $h \in I$ such that $\Delta(x,h)>\epsilon$. In particular, there exists $l \in \mathbb{N}$ such that $\Delta(x,h) > \epsilon + \frac{2}{\ell}$. As $\Delta(x,h) = \lim_{n \to \infty} \Delta_n(x,h)$, we have $\Delta_n(x,h) \geq \epsilon+ \frac{1}{\ell}$ for all $n$ sufficiently large. This proves "$\subseteq$".
Now let $x$ be an element of the right-hand side of $(2)$. Then there exist $\ell ,N \in \mathbb{N}$ such that $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell}$$ for all $n \geq N$. By the definition of $\sup$, we can find $h(n) \in I$ such that $$\Delta_n(x,h(n))> \epsilon + \frac{1}{\ell}. \tag{3}$$ Since $I$ is compact, there exists a convergent subsequence $h_n \to h \in I$. Choose $n=n(x) \geq N$ sufficiently large such that $$|J_n(x)-J(x)| \leq \frac{1}{2k \ell}.$$ This implies $$\frac{|J_n(x)-J(x)|}{h} \leq \frac{1}{2\ell}$$ for all $h \in I$; hence in particular for $h=h_n$. Combining this with $(3)$, we get
$$\begin{align*} \frac{J_n(x+h_n)-J(x)}{h_n} &= \frac{J_n(x+h_n)-J_n(x)}{h_n} + \frac{J_n(x)-J(x)}{h_n}\\ &\geq \epsilon + \frac{1}{\ell} - \frac{1}{2\ell} = \epsilon + \frac{1}{2\ell} > \epsilon. \end{align*} \tag{4}$$
Finally, we note that $J_n(y) \leq J(y)$ implies
$$\epsilon \stackrel{(4)}{<} \frac{J_n(x+h_n)-J(x)}{h_n} \leq \frac{J(x+h_n)-J(x)}{h_n} = \Delta(x,h_n).$$
Consequently, $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h) \geq \Delta(x,h_n) > \epsilon,$$ i.e. $x \in A$.
Step 3: Recall that we actually want to show that $A$ defined in $(1)$ is measurable. By the second part, it suffices to show that sets of the form $$ B := B_{N,n,k,m,\epsilon} := \left\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon \right\}$$ are measurable.
To show this, we use the following elementary result:
Lemma: Let $f: (a,b) \to \mathbb{R}$ be of the form $f = \frac{f_1}{f_2}$ where $f_1$ is increasing and $f_2 \neq 0$ continuous. Then $$\limsup_{\mathbb{Q} \ni q \downarrow x} f(q) \geq f(x)$$ for any $x \in (a,b)$.
Since the mapping $h \mapsto J_n(x+h)-J_n(x)$ is increasing, we may apply the previous result to obtain $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h) = \sup_{(\mathbb{Q} \cap I) \cup \{\frac{1}{m}\}} \Delta_n(x,h) \tag{5}$$ where $I$ is defined as in step 2. (The inequality "$\geq$" holds in any case, the lemma above gives "$\leq$".) Since $\Delta_n(\cdot,h)$ is measurable, this proves that the left-hand side of $(5)$ is measurable and so is $B$.
Best Answer
Using continuity you can prove that $$D^+(F)(x)=\limsup_{h\to 0^+,\, h\in \mathbb Q} \frac{F(x+h)-F(x)}{h}=\lim_{n \to\infty} \sup_{h\in\mathbb Q\cap (0, \frac1n) }\frac{F(x+h)-F(x)}{h}$$ and measurability follows from taking countable supremum and infimum.