Let $B$ be Brownian motion. Consider $g:\mathbb R \to \mathbb R$ that is $C^{2}$ except for some exceptional set $\{ x_{1},...,x_{n}\}\subseteq \mathbb R$. How can we adapt the Ito's formula if $g^{''} \leq M$, for $M > 0$, to show that
$$g(B_{t}) = g(B_{0}) + \int_{0}^{t} g^{\prime}(B_{s})dB_{s}+\frac{1}{2}\int_{0}^{t}g^{\prime\prime}(B_{s})ds \; (*)$$
Due to density arguments we can find $(f_{k})_{k \in \mathbb N} \subseteq C ^{2}$ such that:
$f_{k} \to g, \; (f_{k})^{\prime} \to g^{'},\; \text{ and }(f_{k})^{\prime\prime} \to g^{''} $ uniformly such that $(f_{k})^{\prime\prime}\leq M$
My attempt:
Note that by Ito's formula:
$$f_{k}(B_{t}) = f_{k}(B_{0}) + \int_{0}^{t} f_{k}^{\prime}(B_{s})dB_{s}+\frac{1}{2}\int_{0}^{t}f_{k}^{\prime\prime}(B_{s})ds \; $$
Clearly a.s. for all $t \geq 0$, $\;f_{k}(B_{t})\to g(B_{t})$ and $f_{k}(B_{0})\to g(B_{0})$
Now we need to show that a.s. $\int_{0}^{t} f_{k}^{\prime}(B_{s})dB_{s}\to \int_{0}^{t}g^{\prime}(B_{s})dB_{s}$ in $L^{2}$ ( I thought it would be a.s. be the stochastic integral lacks pathwise interpretability)
$$ \mathbb E \left[\left(\int_{0}^{t} f_{k}^{\prime}(B_{s})dB_{s}- \int_{0}^{t} g^{\prime}(B_{s})dB_{s}\right)^{2}\right]= \mathbb E \left[\int_{0}^{t} \left(f_{k}^{\prime}(B_{s})- g^{\prime}(B_{s})\right)^{2}ds\right]$$ and then use dominated convergence to let it go to zero.
Now onto proving $\int_{0}^{t}f_{k}^{\prime\prime}(B_{s})ds\to\int_{0}^{t}g^{\prime\prime}(B_{s})ds$. Here I am not sure in what convergence sense we should prove (convergence in $L^{2}$ or convergence almost surely, since the Lebesgue-Stieltjes integral does have pathwise interpretation.)
I suggest a.s. convergence:
$\rvert\int_{0}^{t}f_{k}^{\prime\prime}(B_{s})ds-\int_{0}^{t}g^{\prime\prime}(B_{s})ds\lvert \leq \int_{0}^{t} \lvert f_{k}^{\prime\prime}(B_{s})-g^{\prime\prime}(B_{s})\rvert ds $
using the fact that $\lvert f_{k}^{\prime\prime} \lvert ,\; \rvert g^{\prime\prime}\rvert\leq M$, we use dominated convergence again to get the result.
Questions: If the above is correct, we have one term in $(*)$ as a limit in $L^{2}$ while the other terms are a.s. limits. How can we include all of these "different" limits in the equality?
Note that we may witness each of your limits as $L^2$ limits, simply by using the fact that $f \in C^2$, so that uniform convergence on compact sets of $f, f^\prime, f^{\prime \prime}$ can be lifted to $L^2$ convergence. Once you have this, there is no issue interpreting the Itô formula for $g$ in an $L^2$ sense.
As an aside, you should note that the proof of this exercise uses the fact that for any $a \in \mathbb{R}$ the Lebesgue measure of $\{t \geq 0 \, : \, B_t = a\}$ is zero.