I am stuck on trying to show for that a fixed measure space with we have
I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
Here it goes:
Let which means we have
Write where the functions and are measurable functions.
Since the measuable function is integrable, it is almost everywhere finite.
Then WLOG we have
from which it follows that
Similarly for , we obtain
Some cases are simple enough. For example if and then (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, had no finite measure subset the inclusions would be reversed (I don't yet have a proof for this), and the general case would (?!) follow by decomposing your space in this (or some similar) fashion.