Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space $\displaystyle (X,\mathcal F, m)$ is not complete in the sense that

$\displaystyle (X,\mathcal F, m)$ is complete if the all the null sets are measurable.

$\displaystyle \mathcal N = \{A\subset X: \exists B\in \mathcal F \text{ such that } A\subset B, m(B)=0 \}\subset \mathcal F$

So well, lets add the "$\displaystyle (X,\mathcal F, m)$ complete" to the hypothesis.

The case when we have only one sigma algebra is true.

The other one remains yet unclear.