Any Lebesgue probability space is mod. 0 isomorphic to some Polish probability space (with $\sigma$-algebra the completion of Borel algebra, and some Borel probability). I would like to see an example of a Lebesgue probability space which is not isomorphic to such a Polish space.

In other word, I want a Lebesgue probability space $(X,\mathcal A,\mu)$ for which there is no topology $\mathcal T$ such that

1. $\mathcal A$ is the completion of the $\sigma$-algebra generated by $T$ and
2. $X$, endowed with $T$, is a Polish space.

(For me a Lebesgue probability space is a complete probability space $(X,\mathcal A,\mu)$ such that there is some Hausdorff, second countable topology $\mathcal T$ which turns $\mu$ into a Borel probability measure on the completion of $\sigma(\mathcal T)$).

Any Lebesgue probability space is mod. 0 isomorphic to some Polish probability space (with $\sigma$-algebra the completion of Borel algebra, and some Borel probability). I would like to see an example of a Lebesgue probability space which is not isomorphic to such a Polish space.

In other word, I want a Lebesgue probability space $(X,\mathcal A,\mu)$ for which there is no topology $\mathcal T$ such that

1. $\mathcal A$ is the completion of the $\sigma$-algebra generated by $T$ and
2. $X$, endowed with $T$, is a Polish space.

(For me a Lebesgue probability space is a complete probability space $(X,\mathcal A,\mu)$ such that there is some Hausdorff, second countable topology $\mathcal T$ which turns $\mu$ into an inner regular Borel probability measure on the completion of $\sigma(\mathcal T)$).

Edit. I just noticed that I forgot the most important hypothesis that $\mu$ must be inner regular in the definition of Lebesgue probability space. Sorry about that. So in particular $\mu$ is a compact measure. Also this definition is in fact equivalent to Rohlin's original definition.