A sufficient condition for a Borel probability measure to be a Dirac measure

Question

Let $\mu$ be a Borel probability measure on $\mathbb{R}$ and $\mu^k=\mu$ for some natural number $k>1$. Then I have been asked to prove that $\mu=\delta_0$. I sense that I have to somehow exploit the properties of characteristic functions. But I can't proceed. Can I get some help?

Answer 1

Okay so I will go along Kimchi Lover's line, with some more elaborations. I shall be using the property of characteristic functions that states that if $\mu$ and $\nu$ are two Borel probability measures on $\mathbb{R}$ and if $\phi_\mu$ denotes the characteristic function of $\mu$, then $\phi_{\mu*\nu}=\phi_\mu\cdot\phi_\nu$. This enables us to write, apropos of the given problem, $\phi_{\mu^k}=(\phi_\mu)^k=\phi_\mu$. Thus $|\phi_\mu(t)|$ is equal to $0$ or $1$ for each $t\in\mathbb{R}$. Also, $\phi_\mu(0)=1$, and $\phi_\mu$ being a characteristic function is continuous on $\mathbb{R}$, which being connected, we must have $|\phi_\mu(t)|=1$ for each $t\in\mathbb{R}$. Thus $\arg(\phi_\mu)$ is continuous. Also $(\phi_\mu)^k=\phi_\mu$ implies that for each $t\in\mathbb{R}$ $(k-1)\arg(\phi_\mu(t))\equiv 0\,(\mod 2\pi)$. Thus $\arg(\phi_\mu)$ can only assume the discrete values $\frac{2n\pi}{k-1}$, for $n\in\mathbb{Z}$. However $\arg(\phi_\mu)$ being a continuous function on connected set $\mathbb{R}$ must have a connected, and hence in this case, a constant image. However $\phi_\mu(0)=1$ implies that $\arg(\phi_\mu(0))=0$. Thus $\arg(\phi_\mu(t))=0$ for every $t\in\mathbb{R}$. Also we have already shown that $|\phi_\mu(t)|=1$ for each $t\in\mathbb{R}$. These two together imply that $\phi_\mu(t)=1$ for each $t\in\mathbb{R}$. Then $\phi_\mu$ is the characteristic function of the Dirac measure $\delta_0$. Since by uniqueness theorem each characteristic function corresponds to a unique probability measure, we must have $\mu=\delta_0$. Hope it helps.

Answer 2

I understood $\mu^k=\mu$ in a poinwise sense, i.e. for all measurable sets $A$ we should have $\mu(A)^k=\mu(A)$.

For all measurable sets $A$ we have $\mu(A) \in [0;1]$. Now if $$ \mu(A)^k = \mu(A) $$ then we get $\mu(A)\in \{0,1\}$. I.e. every measurable set has either measure $0$ or $1$.

First we have $$ 1 = \mu(\mathbb{R}) = \lim_{n\rightarrow \infty} \mu((-n;n]) $$ Thus, there exists $N\in \mathbb{N}$ such that $\mu((-N; N])=1$. Now we define $(a_0, b_0):=(-N;N)$ and inductively $$ (a_{n+1}, b_{n+1}) := \begin{cases} (a_n, \frac{a_n + b_n}{2}),& \text{if } \mu((a_n, \frac{a_n + b_n}{2}])=1; \\ (\frac{a_n + b_n}{2}, b_n),& \text{if } \mu((\frac{a_n + b_n}{2}; b_n])=1; \end{cases}$$ We define $x_0:=\lim_{n\rightarrow\infty} \frac{a_n + b_n}{2}$ (check that this indeed exists). Then we have $$ \mu(\{ x_0\}) = \lim_{n\rightarrow \infty} \mu((a_{n+1}; b_{n+1}]) = \lim_{n\rightarrow \infty} 1 = 1. $$ Hence, we have shown that our measure is indeed a dirac measure.

Answer 3

Assuming the OP meant to $\mu^k$ to mean the $k$-th convolution product of $\mu$:

If the convolution of $k$ copies of $\mu$, for some $k>1$, is equal to $\mu$, then we know $\phi(t)^k=\phi(t)$ for all real $t$, where $\phi(t)=\int_{\mathbb R} \exp(itx) \mu(dx)$ is the characteristic function of $\mu$. Since characteristic functions are continuous we can see that $|\phi(t)|=1$ for all $t$, and so $\arg(\phi(t))$ is well-defined everywhere, and hence also continuous everywhere. Since $(k-1)\arg\phi(t)\equiv 0 \pmod{2\pi}$ for all $t$, and since $\phi(0)=1$ we see that $\phi(t)=1$ for all $t$. That is, $\phi$ is the characteristic function of the Dirac probability measure $\delta_0$. By the uniqueness-of-characteristic-functions theorem, it is the only such measure.