Let $T:[0,1]\to[0,1]$ be a Lebesgue measurable map which preserves the Lebesgue measure (indicated here by $m(\cdot)$).
Suppose that $m(T(I))= 1$ for every nonempty $T$-invariant interval $I\subset [0,1]$. Does this imply that $T$ is ergodic?
My try: given a measurable set $A$ and $\epsilon>0$, there is an open-interval cover $\{U_n\}_n$ of $A$ such that $|\sum_n m(U_n)-m(A)|<\epsilon$. Then applying the condition to every $U_n$, if thery are invariant, you get the result, but $A$ could be invariant without all $U_n$ being such...
It isn't. Take for instance the rotation by $\frac{1}{2}$, that is,
$$ T(x) = \begin{cases} x+\frac{1}{2}, \ \text{if} \ x < \frac{1}{2}\\ x-\frac{1}{2}, \ \text{if} \ x \geq \frac{1}{2} \end{cases} $$
it can be viewed as a rotation if we identify the endpoints of $[0,1]$ to turn it into a circle. By definition any invariant set must contain full orbits, that is, if $A$ is invariant and $x \in A$, then $T(x)$ is in $A$.
Suppose now that $I$ is an invariant interval and take $c$ its infimum. If $c \geq \frac{1}{2}$, then $T(x) = c - \frac{1}{2} \in I$, which contradicts the definition of infimum. Therefore $c < \frac{1}{2}$ and since $I$ is an interval and $c,c+\frac{1}{2} \in I$, then $\left[c,c+\frac{1}{2}\right] \subset I$. If $c = 0$, then $$\left[0,\frac{1}{2}\right] \subset I \implies [0,1) \subset I \implies m(I) = 1$$ and if $c >0$, then $$\frac{1}{2} \in \left[c,c+\frac{1}{2}\right] \subset I \implies T\left(\frac{1}{2}\right) = 0 \in I$$ which contradicts the fact that $c = \inf I$. Therefore $c = 0$ and $m(I) = 1$, which shows that $m$ is "interval ergodic" with respect to $T$.
But $m$ is not truly ergodic with respect to $T$, indeed, take $J$ a closed small interval contained in $\left(0,\frac{1}{2}\right)$ and $A = J \cup T(J)$ which is an invariant set with $0< m(A) < 1$.