Suppose $\mathcal{A}_1$ and $\mathcal{A}_2$ are $\sigma$-algebras on some set $X$, and let $\nu: \mathcal{A}_1\to \mathbb R$ be a probability measure. If $f:X \to \mathbb R$ is measurable with respect to $\mathcal{A_1}$, we know that it measurable with respect to $\mathcal{A}_2$.
Now, in a measure theory course that I am taking, it contains the remark:
"Since null sets in $\mathcal{A}_1$ are also null sets in $\mathcal{A}_2$, it follows that $Lp(X, \mathcal{A}_1, \nu) \subset Lp(X, \mathcal{A}_2, \nu)$ for every p".
My question: I don't really understand why it is necessary to mention null sets in the above remark. If $f \in Lp(X, \mathcal{A}_1, \nu)$, then by definition it is $\mathcal{A}_1$-measurable, and $\int_X|f|^pd\nu<\infty$. But this implies that $f$ is $\mathcal{A}_2$-measurable (since $\mathcal{A}_1 \subset \mathcal{A}_2)$, and $\int_X|f|^pd\nu<\infty$. Shouldn't this immediately by definition imply that $f \in Lp(X, \mathcal{A}_2, \nu)$? Hence, shouldn't this be sufficient to explain why $Lp(X, \mathcal{A}_1, \nu) \subset Lp(X, \mathcal{A}_2, \nu)$ for every $p$? I do not understand why it is necessary to mention null sets here.