Convolution of a probability measure with itself is the same measure Say you have a probability measure $\mu$ on $\mathbb{R}$.
Suppose $\mu*\mu = \mu$. Then prove that $\mu = \delta_0$ i.e $\mu$ is the Dirac measure at zero. 
I know this can be proved using uniqueness characteristic functions. However, I was wondering if there is a way to prove this purely using Measure Theoretic Probability. 
I've tried to prove that $\mu(\{0\}) = 1$ since this should be sufficient but I've not been able to. Despite that, I can't help but feel there is some trick or technique which I've missed.
 A: It is easier to argue using random variables. Let $X$ and $Y$ be i.i.d. with distribution $\mu$. Note that $X+Y >0$ implies either $X>0$ or $Y>0$. Hence $P(X+Y >0) \leq P(X>0)P(Y>0)$. But $X+Y$ also has distribution $\mu$. Hence $P(X>0) \leq (P(X>0))^{2}$. This gives $P(X>0)=0$ or $1$. 
Suppose $P(X>0)=1$. $Ee^{-X}=Ee^{-X-Y}=Ee^{-X}Ee^{-Y}=(Ee^{-X})^{2}$, so $Ee^{-X}=0$ or $1$. But this can never be $0$, so $Ee^{-X}=1$. But then $1-e^{-X}$ is a positive random variable with mean $0$. Hence $1-e^{-X}=0$ almost surely or $X=0$ almost surely. The case when $P(X<0)=1$ is similar. 
A: The space of finite measures on $\Bbb{R}$ with the convolution(as multiplication) and the norm $||\mu||=|\mu|(\Bbb{R})$ is a Banach Algebra.
You can prove this using Fubini's Theorem.(The inequality $||\mu \ast\nu|| \leq||\mu||||\nu||$)
This algebra has a unit element which is the Dirac measure  $\delta_0$(again this is proved by Fubini's theorem and the definition of convolution of measures)
The unit element with respect to the multiplication of a Banach algebra is unique.
So in your case $\mu=\delta_0$
