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I'm working on a paper in which I have to prove that following measure is $T$ invariant.
Notice that $T$ is defined the way like this :

\begin{align*} T: [0,1] \longrightarrow [0,1] \\ T(x) = \begin{cases} x+ \dfrac{1}{2} & x \in [0,\dfrac{1}{2}]\\ 2x-1 & x \in (\dfrac{1}{2} , 1] \end{cases} \end{align*}
I'm going to prove that $\mu$ is $T$ invariant
\begin{align*} \mu = \dfrac{\nu + \nu T}{2} \end{align*}
in which $\nu$ is a $ T^{2} $ invariant measure with density $h$ with respect to lebesuge measure $m$.

I tried alot :

As you know we have to prove for any $B$ in $\mathcal{B}[0,1] $ we have : \begin{align*} \mu(B) = \mu(T^{-1}B) \end{align*}
So :
\begin{align*} \mu(B) = \dfrac{\nu(B) + \nu T(B)}{2} \end{align*} And we have : \begin{align*} \mu(T^{-1}B) = \dfrac{\nu T^{-1}(B) + \nu(B)}{2} \end{align*}
So we have to prove that :
\begin{align*} \nu T (B) = \nu T^{-1}(B) \end{align*}
We know that $\nu T^{-1}(B) = \int_{T^{-1} (B)} h dm$
I computed this :
\begin{align*} \nu T^{-1}(B) = \dfrac{1}{2} h(\dfrac{x+1}{2}) + h(x - \dfrac{1}{2}) 1 _{[\dfrac{1}{2},1]}(x) \end{align*}

I don't know how compute $\nu T(B)$ and when do we use from the fact that $\nu$ is $T^{2}$ invariant ?
Can we say that since $ \nu $ is $ T^{2} $ invariant so : \begin{align*} \nu(B) = \nu(T^{-2}(B)) \end{align*} So we have that : \begin{align*} \nu( T T^{-2}(B)) = \nu (TB) \end{align*}

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Since $\nu T^{-2}=\nu$, we have $$\nu(T(B))=\nu(T^{-2}(T(B)))=\nu(T^{-1}(B)),$$ and thus $$\mu(T^{-1}(B))=\frac{1}{2}\left[\nu(T^{-1}(B))+\nu(B)\right]=\frac{1}{2}\left[\nu(T(B))+\nu(B)\right]=\mu(B).$$

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Think of $\mu$ as being the average of $\nu$ and $\nu T$. Now if $\nu$ is $T^2$ invariant, this means that $\nu$ and $\nu T^2$ is the same measure and you should be able to conclude directly that $\mu$ is $T$-invariant. (You average along the $T$ orbit using the periodic behaviour of $\nu$.)

Bonus question: If you started with a measure say $\kappa$ which is $T^n$ invariant for some $n$ a fixed natural number. How would you construct a $T$-invariant measure from $\kappa$?

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