There are many criteria for uniform convergence of continuous functions, such as Stone-Weirestrass etc… However are there any known results guaranteing that a sequence of continuous functions $\{f_n\}$ in $C(R)$ converges pointwise to some discontinuous function?
Criteria for Pointwise Convergence of Continuous Functions
calculusconvergence-divergencefunctional-analysisreal-analysis
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I think your argument works, the only thing that isn't completely justified is why you can conclude that since $\hat{f}_n$ converges uniformly to $f$, so does $f_n$. And where do you use that there are only a finite number of points of discontinuity (the example below shows that this assumption is necessary). You can complete the argument as follows; Given $\epsilon>0$ there exists $\hat{N}$ and $\delta>0$ such that $$ n>\hat{N} \ \Rightarrow \ \forall x \ : \ |\hat{f}_n(x)-f(x)|<\epsilon$$ Label the points of discontinuity $\{y_1,\ldots,y_k\}$. By pointwise convergence we know that for each point of discontinuity $y_j$, there exists and $N_j$ such that $$ n>N_j \ \Rightarrow \ |f_n(x)-\hat{f}_n(x)|<\epsilon$$ But now taking $N=\max(\hat{N},N_1,\ldots,N_k)$ should give $$ n>N \ \Rightarrow \ |f_n(x)-f(x)|<2\epsilon$$
EDIT:The following answers a misunderstood version of the question, but since reference to the example is made above, I will leave it up. Missed the assumption that points of discontinuity are assumed independent of $n$ and finite in number.
I am not convinced that this is necessarily true. How about the sequence of functions $$ f_n(x) = \begin{cases} 0 & \text{ if } x\in ]-\infty,-1/n[\cup [0;\infty[\\ 1/2+1/n & \text{ if }x\in [-1/n;0[ \\ \end{cases}$$ seems to me like this sequence of functions will converge pointwise to $0$ and be stictly decreasing, but it is clearly not uniformly convergent, since $$ \forall n : \ \max(\{|f_n(-\delta) - f_n(0)| \ \big| \ \delta \in [-1;0]\}) >1/2$$
The trick here is to use bump functions. Specifically, we are using the fact that if $K\subseteq U$ with $K$ compact and $U$ open, and $c_1$ and $c_2$ are constants, then there is a smooth (i.e., in $C^\infty(\mathbb R^d)$) function equaling $c_1$ on $K$ and $c_2$ outside of $U$, and with values between the two constants everywhere.
Let $h_n$ be a sequence of smooth functions converging pointwise a.e. to $g$, as you have described. Notice also that if $\psi_n$ are smooth bump functions equaling $1$ on the balls of radius $n$ about the origin, and then decaying to $0$, then $h_n\psi_n$ also converges pointwise a.e. to $g$, so henceforth we assume WLOG that each $h_n$ is compactly supported.
For each $n$, let $U_n=\{x\in\mathbb R^n\mid |h_n(x)|> 1+\frac{1}{n}\}$, and let $K_n=\{x\in\mathbb R^n\mid |h_n(x)|\geq 1+\frac{2}{n}\}$. Then $K_n\subseteq U_n$ with $K_n$ compact and $U_n$ open, so let $\phi_n$ be a smooth bump function bounded by $\frac{1}{1+\frac{2}{n}}$, equal to $\frac{1}{1+\frac{2}{n}}$ outside of $U_n$ and equal to $0$ on $K_n$. Then each $h_n\phi_n$ is smooth and compactly supported (hence all derivatives bounded) and bounded by $1$ in absolute value everywhere, and $h_n\phi_n\rightarrow g$ pointwise a.e. (since at any $x$ where $h_n(x)$ converges to $g(x)$, eventually $x$ will lie outside of every $U_n$).
Best Answer
Generally we can use well known fact, that for convergence, for fixed $x$, $f_n(x)$ should be Cauchy sequence
For exact condition which gives continuity for sequence continuous functions limit let's consider introduced by C.Arzela so called quasiuniform convergence: We say, that on some $[a,b]$ segment sequence of continuous functions $f_n(x)$ quasiuniformly converged to continuous function $f(x)$, if for any $\forall \epsilon$ and any $N$ segment $[a,b]$ can be covered by finite amount intervals $(a_1,b_1),(a_1,b_1), \cdots, (a_i,b_i), \cdots, (a_k,b_k)$ and they can be assigned to numbers $n_1,n_2, \cdots, n_i, \cdots, n_k$ $(>N)$ that for every $x$ from $(a_i,b_i)$ performed simultaneously $|f(x)-f_{n_i}(x)|< \epsilon$.
Now using this concept Arzela proved theorem:
Suppose sequence of continuous functions $f_n(x)$ pointwise converged on $[a,b]$ segment to function $f(x)$. Then for $f(x)$ continuity is necessary and sufficient, that $f_n(x)$ converged quasiuniformly.