trivial map on $\sigma-$algebra $\mod{}0$ is trivial Hi everyone!
I am currently studying the basic theory of  measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost... 

Question: Suppose $T$ is an invertible
  measure-preserving map of standard probability measure
  space $(X,\mu)$. Suppose that $TA=A$
  for all measurable subsets
  $A\subset{}X$, where the equality is
  up to sets of measure $0$. Prove that
  the set of those $x$ where $Tx\neq{}x$
  has measure $0$.

 A: If $X$ is a standard probability space then we may assume it to be the disjoint union of an interval with Lebesgue measure and a countable set of atoms. If $p$ is an atom, then by assumption, $T(p) = p$, since $p$ has positive measure. So none of the atoms can be "bad" points. So we may assume that there are no atoms, so that $X=[0,1]$ and the measure is Lebesgue measure. Now consider all subintervals $I$ of $[0,1]$ with rational endpoints and gather all points in $TI$ which are outside $I$. We get a countable union of measure zero sets, hence a measure zero set. Denote it by $Z$. Now let $x \notin Z$. This implies that $Tx$ belongs to arbitrarily small intervals around $x$ and is thus equal to $x$. The desired result follows.
A: not an answer to the revised question
But Das Curious asks for the counterexample (to the original question) when the countability assumption is omitted.  Here it is.
Let $X = [0,1)^{[0,1]}$, that is: the set of all functions $x \colon [0,1] \to [0,1)$.  Let $\lambda$ be Lebesgue measure on $[0,1)$.  Let $\Lambda$ be the product measure on $X$ with the product $\sigma$-algebra $\mathcal F$.  In particular: if $A \in \mathcal F$, then there is a countable $J \subseteq [0,1]$ such that $A$ "depends only on" $J$ in the sense that: for $x,y\in X$ with $x(t)=y(t)$ for all $t \in J$, then either $x,y$ both belong to $A$ or both belong to $X \setminus A$.
Let $\theta$ be the "irrational rotation" $\theta \colon [0,1) \to [0,1)$ defined by $t \mapsto t+\sqrt{2} \pmod{1}$.  So $\theta$ is a measurable, measure-preserving bijection (with no fixed point) of $([0,1),\lambda)$ onto itself.  For a fixed $s \in [0,1]$, define $\theta_s \colon X \to X$ by "rotation of coordinate $s$", namely
$\theta_s(x)(s) = \theta(x(s))$ and $\theta_s(x)(t) = x(t)$ for all $t \ne s$.  So $\theta_s$ is a measurable, measure-preserving bijection of $(X,\Lambda)$ onto itself.
For $s \in [0,1)$ let $Q_s = \{x \in X\;|\; x(1)=s\}$.  Thus $(Q_s)_{s \in [0,1)}$ is a family of subsets of $X$; the family has cardinal of the continuum, so it is indexed by $[0,1)$; each $Q_s$ is a measurable set with $\Lambda(Q_s)=0$; the sets $Q_s$ are pairwise disjoint; but the union $\bigcup_{s} Q_s = X$ is the whole space.    
Now we can define $T \colon X \to X$.  Let
$$
T(x)(t) = \theta(x(t))\qquad\text{if } x \in Q_{t}
\\
T(x)(t) = x(t)\qquad\text{if }x \notin Q_t
$$
Now each $x$ belongs to exactly one $Q_t$, so $x$ and $T(x)$ agree, except on a single coordinate, where they disagree.  Note $T(x) \ne x$ for all $x$.  Because $1 \notin [0,1)$, the map $T$ maps each $Q_s$ into itself.  Thus $T$ is a bijection of $X$ onto itself:  the inverse map has the same definition using $\theta^{-1}$ in place of $\theta$.
We claim that the map $T$ is measurable.  Let $A \in \mathcal F$.  We claim $T^{-1}(A) \in \mathcal F$.  There is a countable set $J \subseteq [0,1)$ such that $A$ depends only on $J$.  For $s \in J$ let $A_s := A \cap Q_s$.  Let $B := A \setminus \bigcup_{s \in J} Q_s$.  Now $A$ is the countable union
$$
A = B \cup \bigcup_{s \in J} A_s
$$
where $B$ and all $A_s$ belong to $\mathcal F$.
The inverse image $T^{-1}(A)$ is the countable union
$$
T^{-1}(A) = T^{-1}(B) \cup \bigcup_{s \in J} T^{-1}(A_s) .
$$
Now by the definition of $J$, we have $T^{-1}(B)=B$.  Each $\theta_s$ is a measurable function, and
$T^{-1}(A_s) = \theta_s^{-1}(A_s)$, so each $T^{-1}(A_s) \in \cal F$.  Therefore $T^{-1}(A)$ is written as a countable union of sets from $\mathcal F$.
We claim $A = T(A)$ up to null sets for any $A \in \mathcal F$.
Let $J$ be as before.
If $x$ is not in the null set $\bigcup_{s \in J} Q_s$, then either $x,T(x)$ are both in $A$ or both in $X \setminus A$.  So $A$ and $T(A)$ agree except for adding and subtracting parts of that null set.
