$\phi$ invariant Borel probability measure

Question

Given $X$ and $\phi:[0,1]\rightarrow[0,1]$

$\phi(x)= \begin{cases} \frac{1}{2}x & x\neq0\\ 1&x=0\end{cases}$

I've to show that there can't be a Borel-probability measure $\mu$ which is $\phi$ invariant (i.e. $\mu(\phi^{-1}(A))=\mu(A)$ $\forall A\in \mathcal{B}(X)$)

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My first idea was proving it by contradiction. But without success.

Answer 1

Suppose such probability measure $P$ on $[0, 1]$ exists. Suppose $F(x)$ is its CDF. By our supposition:

$F(x) = \begin{cases} 1 - P(\{0\} ) & \quad x > \frac{1}{2}\\ F(2x) - P(\{0\}) & \quad x \leq \frac{1}{2} \end{cases}$

If $P(\{0\}) > 0$, then $\exists x > 0 \text{ such that } F(x) < 0$, which is impossible.

If $P(\{0\}) = 0$, then $\forall x > 0 \text{ } F(x) = 1$, which is also impossible.

Thus we receive a contradiction.