The Ergodicity of Baker's transformation How to prove that Baker's transformation is ergodic with respect to Lebesgue measure directly without showing that it is isomorphic to bernoulli shift ? 
 A: There is a martingale proof of ergodicity (I suppose that the mentioned transformation is $\theta:[0,1)\to [0,1),$ $\theta(x)=\{2x\}$). Let $A$ be a $\theta-$invariant Borel subset of  $[0,1),$ i.e.
$$
A=\theta^{-1}(A)=\bigg(\frac{1}{2}A\bigg)\cup \bigg(\frac{1}{2}+\frac{1}{2}A\bigg).
$$
By induction, for any $n\geq 1$ we can write $A$ as a disjoint union
$$
A=\bigcup^{2^n-1}_{k=0}\bigg(\frac{k}{2^n}+\frac{1}{2^n}A\bigg).
$$
Let $\mathcal{F}_n$ be a $\sigma-$field generated by partition
$$
[0,1)=\bigcup^{2^n-1}_{k=0}\bigg[\frac{k}{2^n},\frac{k+1}{2^n}\bigg).
$$
Then $\bigcup_{n\geq 1}\mathcal{F}_n$ generates the Borel $\sigma-$field on $[0,1).$ $P$ will denote the Lebesgue measure on $[0,1).$
The sequence $X_n=P(A|\mathcal{F}_n)$ is a bounded martingale, which converges a.s. to $1_A.$ It remains to check that $X_n=P(A)$ a.s. and deduce that a.s. $P(A)=1_A\in \{0,1\}.$ But 
 $$
 X_n=P(A|\mathcal{F}_n)=\sum^{2^n-1}_{l=0}2^n P\bigg(A\cap \bigg[\frac{l}{2^n},\frac{l+1}{2^n}\bigg) \bigg)1_{[\frac{l}{2^n},\frac{l+1}{2^n})};
 $$
$$
  P\bigg(A\cap \bigg[\frac{l}{2^n},\frac{l+1}{2^n}\bigg) \bigg)=\sum^{2^n-1}_{k=0}P\bigg(\bigg(\frac{k}{2^n}+\frac{1}{2^n}A\bigg)\cap [\frac{l}{2^n},\frac{l+1}{2^n})\bigg)=
 $$
$$
 =P\bigg(\bigg(\frac{l}{2^n}+\frac{1}{2^n}A\bigg)\cap [\frac{l}{2^n},\frac{l+1}{2^n})\bigg)=P\bigg(\frac{l}{2^n}+\frac{1}{2^n}A\bigg)=\frac{P(A)}{2^n},
 $$
 and $X_n=P(A)$ a.s.
P.S. If one does not like martingales, the only fact that is really needed here is that for any Borel set $A\subset [0,1)$
$$
\int^1_0 \bigg|\sum^{2^n-1}_{l=0}2^n P\bigg(A\cap \bigg[\frac{l}{2^n},\frac{l+1}{2^n}\bigg) \bigg)1_{[\frac{l}{2^n},\frac{l+1}{2^n})}(x)-1_A(x)\bigg|dx\to 0, \ n\to\infty.
$$
This can be proved by means of real analysis only. 
