Well-definedness of inclusion of $L^{p}$-spaces.

Question

Let $(X,\mu)$ be a finite measure space and suppose that $p,q\in[0,\infty)$ such that $p\geq q$. Recall that

$$L^{p}(X,\mu):=\mathcal{L}^{p}(X,\mu)/N(X,\mu),$$

where $\mathcal{L}^{p}(X,\mu)$ is the space of $p$-integrable complex-valued functions on $X$ and

$$N(X,\mu)=\{(f\colon X\to\mathbb{C}):f \ \text{measurable}, \ f=0 \ \text{$\mu$-a.e}\}.$$

I have seen that $\mathcal{L}^{p}(X,\mu)\subset\mathcal{L}^{q}(X,\mu)$.

• But does this also imply that $L^{p}(X,\mu)\subset L^{q}(X,\mu)$? I mean, since $L^{p}(X,\mu)$ and $L^{q}(X,\mu)$ consist of equivalence classes, is the latter inclusion actually well-defined (in the set-theoretic sense)?
• Is the above inclusion also true for $q=\infty$?

I know that there are several posts on Stack-Exchange about this inclusion, but I can't seem to find any posts + answers that properly discuss the fact that $L^{p}(X,\mu)$ and $L^{p}(X,\mu)$ consist of equivalence classes.

Answer 1

The inclusion you are asking about does also hold. Think about quotient spaces of vector spaces. If we have two subspaces $U,V$of a vector space $X$ and a closed subspace $S$ of both $U,V$ we can consider the following. $$\pi_{V}:V\rightarrow V/S $$ $$\pi_{U}:U\rightarrow U/S$$ In our case we also have the fact that $U \subset V$ so you can check that the inclusion is preserved after we quotient.

Hope this helps.

Answer 2

Yes, there is absolutely no issue with this because $N (X,\mu) $ is the same regardless of $p,q $.

So you have an inclusion $A\subset B $ of vector spaces, and a common subspace $N $. An element of $A/N $ is $a+N $ with $a\in A $. As $a\in B $, the set $a+N $ is also the class of $a $ in $B/N $.