Let $T : (\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda^1) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda^1)$ be measurable. I claim that $T$ is measure preserving on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda^1)$ if and only if $$ \lambda^1(G) =\lambda^1(T^{-1}(G)) \ \ \ \ (\ast) $$ for all open $G\subseteq \mathbb{R}$. We denote the collection of open sets by $\mathcal{O}(\mathbb{R})$
The way I am trying to show this is by the following;
Clearly if $T$ is measure preserving then $(\ast)$ holds by definition.
Now suppose that $(\ast)$ holds, Then consider the system
$$ \Sigma = \{A \subset \mathbb{R}|\lambda^1(A) =\lambda^1(T^{-1}(A))\} $$
Then clearly $\mathcal{O}(\mathbb{R})\subseteq \Sigma$. So if we can show that $\Sigma$ is $\sigma$-algebra then we have $$ \mathcal{B}(\mathbb{R})= \sigma(\mathcal{O}(\mathbb{R}))\subseteq \sigma(\Sigma) = \Sigma. $$
Now showing that $\Sigma$ is a Dynkin System is easy enough, but now I am trying( and failing) to show that $\Sigma$ is closed under finite intersections. If I can show that $\Sigma$ is closed under finite intersections, then it will be $\sigma$-algebra as needed.
Cheers in advance for any help.
You cannot prove directly that $\Sigma$ is a sigma algebra. The collection $\mathcal U$ of open sets is a $\pi-$ system contained in the Dynkin system $\Sigma$ and this implies that the sigma algebra generated by open sets is contained in $\Sigma$.