Let $f : [0,1] \to [0,1]$, $$ f(x) = \frac{x}{2}, \;\; 0<x\leq 1 \;\; and \;\; f(x) = 1, x=0.$$ I'm trying to show that $f$ has no invariant Borel probability measure.
My work: Suppose that there is an invariant Borel probability measure $\mu$. Then $\mu(A) = \mu(f^{-1}(A)), \forall A \in \mathcal{B}([0,1]).$
$[\frac{1}{2}, 1] \in \mathcal{B}([0,1])$
$f^{-1}([\frac{1}{2}, 1]) = \{x \in [0,1]: f(x) \in [\frac{1}{2}, 1]\} =\{0, 1\}$. As $\mu$ is invariant, then we have $\mu(f^{-1}[\frac{1}{2}, 1]) = \mu([\frac{1}{2}, 1])$.
Hence $\mu([\frac{1}{2}, 1]) = \mu(\{0, 1\})$.
Can we say that $\mu([\frac{1}{2}, 1]) \neq 0$ and $\mu(\{0, 1\}) = 0$? (And so will result a contradiction.)
Thank you!
What you did is correct, but I think it's not enough. Assume that $\mu$ is an invariant probability measure for $f$.
You have $$\tag{*}\mu(\{0\})=\mu(f^{-1}(\{1\})=\mu(\{1\}). $$ Since $f^{-1}((1/2,1))=\emptyset$, we get $$ \mu((0,1])=\mu(f^{-1}((0,1/2]))=\mu((0,1/2]) $$
You have $$ 0=\mu(\emptyset)=\mu(f^{-1}(\{0\}))=\mu(\{0\}), $$ so $$\mu([0,1])=\mu(\{0\}\cup(0,1])=\mu((0,1]).$$ Also, $$ \mu\left(\,\left(0,\frac1{2^n}\right]\,\right) =\mu\left(f^{-1}\left(0,\frac1{2^n}\right]\,\right) =\mu\left(\,\left(0,\frac2{2^n}\right]\,\right) =\mu\left(\,\left(0,\frac1{2^{n-1}}\right]\,\right). $$ It follows that $$1=\mu([0,1])=\mu([0,1/2])=\mu([0,1/2^n])\xrightarrow[n\to\infty]{}\mu(\{0\})=0$$ by continuity of the measure. The contradiction shows that $\mu$ cannot exist.