I am stuck on trying to show for that a fixed measure space with we have

.

Cheers!

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- May 11th 2010, 10:49 PMwilly0625Lp spaces set inclusion
I am stuck on trying to show for that a fixed measure space with we have

.

Cheers! - May 12th 2010, 01:55 PMwilly0625
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

- May 14th 2010, 07:44 PMJose27
I don't like your argument, you can't say (what if is negative), and the part where you took integrals is fishy at best.

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. - May 15th 2010, 12:10 AMOpalg