What is the total variation of a dirac delta function $\delta(x)$? What is the total variation of a dirac delta function $\delta(x)$? My guess is that it is something like $\infty$. If not defined, what would be the best way to define?
 A: Let $(f_1,f_2,f_3,\ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$\lim_{n\to\infty}\int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$\lim_{n\to\infty}\mathrm{TV}(f_n)=\infty.$$This is one way to interpret your question, and then the answer is indeed $\infty$.
A: To clarify this, lets agree on the definitions of which we're using.  Let $(X,\Sigma)$ be a measurable space.  
For any measure $\mu$ on $(X,\Sigma)$ define, the upper and lower variation (respectively) as follows
$$
\overline{\mathrm{W}}(\mu,E)=\sup\left\{\mu(A)\mid A\in\Sigma\text{ and }A\subset E \right\}\qquad\forall E\in\Sigma\\
\underline{\mathrm{W}}(\mu,E)=\inf\left\{\mu(A)\mid A\in\Sigma\text{ and }A\subset E \right\}\qquad\forall E\in\Sigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
\|\mu\|_{TV}\triangleq \sup_{E \in \Sigma} \left(
\overline{\mathrm{W}}(\mu,E) + \left| \underline{\mathrm{W}}(\mu,E)\right|
\right)
= \overline{\mathrm{W}}(\mu,X) + \left| \underline{\mathrm{W}}(\mu,X)\right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$\delta_B(A)\triangleq \begin{cases}
1 : & A \cap B\neq\emptyset\\
0 : & \mbox{else}\\
\end{cases},
$$
then
$$
\overline{\mathrm{W}}(\mu,X)=\sup\left\{\delta(A)\mid A\in\Sigma\text{ and }A\subset E \right\}=
\delta(B)=1.  
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$.  Therefore, 
$$
\|\delta_B\|_{TV}=1<\infty.
$$
Definition 2: Physics
Fix some $x \in X$ and define the Dirac delta (generalized) function as
$$
\delta_x(A)\triangleq \begin{cases}
\infty : & x \in A\\
0 : & x \not\in A
\end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $\infty$, and is maximized by $\{x\}$.  Likewise, the lower variation is minimized to value $0$ (if $\# X>1$).  Hence
$$
\|\delta\|_{TV}=\infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.  
