Cut out a ball of radius $\tfrac1n$ around each discontinuity $x_k$ (assume $n$ is large enough that these balls are all disjoint). Then just define $f_n$ to agree with $f$ on the complement of the balls. On the ball around $x_k$, define $f_n$ to be linear so that it "spans the gap", i.e., connects the points $(x_k-\tfrac1n,f(x_k-\tfrac1n))$ and $(x_k+\tfrac1n,f(x_k+\tfrac1n))$.
It may be necessary to make two segments if the the linear part doesn't happen to agree with $f$ at $x_k$. But that's easy: first span the gap between $(x_k-\tfrac1n,f(x_k-\tfrac1n))$ and $(x_k,f(x_k))$, then span the gap between $(x_k,f(x_k))$ and $(x_k+\tfrac1n,f(x_k+\tfrac1n))$. This ensures that $f_n(x_k)=f(x_k)$ for all $n$ and $k$.
You need to be careful about how exactly your subsequences are constructed.
Take a sequence $r_k\nearrow 1$ as $k\to\infty$, and let $B_k := B(0,r_k)$.
Let us start with $B_1$. By the assumptions there is a subsequence $f_{\sigma(1,1)}, f_{\sigma(1,2)}, f_{\sigma(1,3)},\dotsc$ which converges pointwise a.e. in $B_1$. Let $Z_1\subset B_1$ be the measure zero set where the pointwise convergence fails, so $(f_{\sigma(1,n)})$ converges pointwise everywhere in $B_1\setminus Z_1$ as $n\to\infty$.
Now, for $B_2$, we let $(f_{\sigma(2,n)})$ be a further subsequence of $(f_{\sigma(1,n)})$ that converges pointwise a.e. in $B_2$. This is possible since $(f_{\sigma(1,n)})$ certainly converges in $L_{\rm loc}^p$ as well. Continuing in this manner, we let $(f_{\sigma(k,n)})$ be a subsequence of the $(f_n)$ with the properties that
- $(f_{\sigma(k,n)})$ is a subsequence of $(f_{\sigma(k-1,n)})$
- $(f_{\sigma(k,n)})$ converges pointwise a.e. in $B_k$, with exceptional measure zero set $Z_k\subset B_k$
Let's arrange everything into a grid:
\begin{equation*}
\begin{array}{cccc}
f_{\sigma(1,1)} & f_{\sigma(1,2)} & f_{\sigma(1,3)} &\cdots \\
f_{\sigma(2,1)} &f_{\sigma(2,2)} & f_{\sigma(2,3)} & \cdots \\
f_{\sigma(3,1)} &f_{\sigma(3,2)} & f_{\sigma(3,3)} & \cdots \\
\vdots & \vdots & \vdots & \ddots
\end{array}
\end{equation*}
It's then not hard to see that the diagonal sequence $g_n = f_{\sigma(n,n)}$ is such that $(g_n)_{n\geq k}$ is a subsequence of $(f_{\sigma(k,n)})_{n\geq 1}$.
Let $Z = \bigcup_i Z_i$, and note that $Z$ is measure zero as the countable union of measure zero sets. Let $x\in B\setminus Z$. Then $x\in B_k\setminus Z_k$ for some $k$. For this $k$, we see that $g_n(x)\to f(x)$ as $n\to\infty$ since, for $n\geq k$, $g_n(x) = f_{\sigma(n,n)}(x)$ is a subsequence of $(f_{\sigma(k,n)}(x))$, the latter of which has limit $f(x)$ by construction. Thus the $(g_n)$ converge pointwise everywhere in $B\setminus Z$, and is thus the desired subsequence of the original $(f_n)$.
Best Answer
Yes, this is possible. Set $f(x) = 0$ and $f_1(x) = (1+x^2)^{-1}$. Then set $$ f_n(x) = n f_1(nx) = \frac{n}{1 + n^2x^2} $$ Clearly $f_n(x) \to 0$ for $x \ne 0$ but $f_n(0)$ does not converge.
If you want the same phenomenon with a $C^k$ function $f$, just use $f(x) + f_n(x)$ instead of $f_n$.