Pointwise convergence of functions with jumps to a smooth function possible? Can a sequence of functions having jump discontinuities converge to a smooth function, pointwise?
If not to a smooth function, converge to some other function which does not have jumps. I just want jumps to disappear!
 A: Yep, consider the following sequence of functions:
\begin{align*}
   f_n(x) = \left\{
     \begin{array}{lr}
       \frac{1}{n} & : x \leq 0\\
       0 & : x > 0
     \end{array}
   \right.
\end{align*}
Each function is discontinuous at $x=0$ (the limit does not exist there), but as $n\to\infty$ the sequence will converge pointwise to $f\equiv 0$, which is smooth. It's the smoothest function I can think of really.
As a slightly more exciting example, consider the sequence:
\begin{align*}
   g_n(x) = \left\{
     \begin{array}{lr}
       \frac{n+1}{n} & : x \in\mathbb{Q}\\
       1 & : x\not\in\mathbb{Q}
     \end{array}
   \right.
\end{align*}
Each function in this sequence is continuous nowhere. In fact, $g_n$ isn't even Riemann integrable. This arises from the fact that in any nonempty subset $S$ of the reals, there will be a rational and irrational number. So the supremum and infimum of the image of $g_n$ for any open interval in $\mathbb{R}$ will be $\frac{n+1}{n}$ and 1 respectively. Still, as $n\to\infty$, $g_n\to g\equiv 1$ pointwise, which is a smooth function.
A: As a contrast to Alex Siryj's examples where the jumps get smaller, consider 
\begin{align*}
   h_n(x) = \left\{
     \begin{array}{lcc}
       0 & : &x \leq 0\\
       n^2 & : &0 \lt x \leq \frac1n\\
       0 & : &\frac1n \lt x
     \end{array}
   \right.
\end{align*}
So there are two jumps getting bigger as $n$ increases and $\int_{-\infty}^\infty h_n(x)\, dx =n$ also increases, but there is pointwise convergence to $h(x)=0$ as $n$ increases and $\int_{-\infty}^\infty h(x)\, dx =0$.
