How to expalin the pathologie of this zero distribution(or measure). Assume $\delta$ is the zero Dirac distribution (measure) on $\Bbb R$. namely,
$$(\delta,f)= \int_\Bbb R f\delta(dx) =f(0)$$
We know if $T\in \mathcal {D'}(\Bbb R)$ is a distribution then for every $ \phi \in C^\infty(\Bbb R)$ we have that  $\phi T$ is also a distribution defines by
$$(\phi T, f) = ( T, \phi f)$$ 
Moreover, $$(\phi T)'= \phi'T+\phi T' $$
My problem. If we consiser the particular case $\phi= x$ or $\phi=x^2$ and $T= \delta$
we have, $$(x\delta, f)= (xf)(0)=0~~~\forall ~~f$$

Question: does this means that $\color{red}{x\delta \equiv 0}$ in $ \mathcal {D'}(\Bbb R)$?



*

*If yes how to explain the fact that $$\color{blue}{0 =(x\delta)' = \delta +x\delta'}$$

*If no what is $\color{red}{x\delta }$ as a distribution?


Patently if one consider the measure $$\mu(dx) = x^2 \delta(dx)$$
then it happens that $$\int f\mu(dx) = 0~~~\forall ~~f$$
Why is $\mu$ a trivial measure.?
 A: Yes, $x\delta=0$. This is obvious from the definitions: $$(x\delta)(f)=\delta(xf(x))=0f(0)=0.$$
Yes, it follows that $\delta + x\delta'=0$. You ask how to explain this. What's to explain? You've proved it's true; why shouldn't it be true?
It's also obvious from the definitions that $x\delta' = -\delta$:
$$(x\delta')(f)=\delta'(xf(x))=-\delta((xf(x))')=-\delta(xf'(x)+f(x))=-f(0). $$
A: As you defined it $\mu \ll \delta$ and $\frac{d \mu}{d \delta}(x) = x^{2}$.  That's fine, but $x^{2} = 0$ $\delta$-almost everywhere.  Thus, we can say $\frac{d \mu}{d \delta}(x) = 0$ $\delta$-a.e.  Therefore, $\mu = 0$.  
The equation $0 = \delta + x \delta'$ does indeed seem strange.  Then again, $\langle \delta', f \rangle = - f'(0)$ (by the definition of distributional derivative) so
\begin{align*}
\forall \psi \in C_{c}^{\infty}(\mathbb{R}) \quad \langle x \delta', \psi \rangle &= \langle \delta', x \psi \rangle \\
&= -(x \psi)'(0) \\
&= - 0 \cdot \psi'(0) - (1) \cdot \psi(0) \\
&= - \psi(0) \\
&= -\langle \delta, \psi \rangle.
\end{align*}
This proves $\delta + x \delta' = 0$ directly.
