I believe the title of the post is almost self explanatory but either way I would like to elaborate a little more on the topic.
Let us fix a Lebesgue-measurable function $p\colon \Omega \to [1,\infty]$. The question I am studying is quite simple and can be stated as follows:
What conclusions can we retrieve about $p(\cdot)$ if we know that $p_+ = \text{ess} \sup_{x \in \Omega} p(x) < \infty$ or $p_- = \text{ess} \inf_{x \in \Omega} p(x) = \infty? $
Here, when I say "conclusions", I am mainly interested in boundedness properties of the function $p(\cdot)$ and it is also important to say that I am studying each one of the possibilities $p_+ < \infty$ or $p_- = \infty,$ individually.
My thoughts. I dealt with the case $p_+ < \infty$ quite easily. I was able to prove that
$$ p_+ < \infty \implies p(\cdot) \text{ is essentially bounded.} $$
If someone whises, I could post further details on this implication.
Now, on the other hand, I am having quite some trouble with the case $p_- = \infty.$ Any hint to proceed developing this case would be kindly appreciatted.
Thanks for any help in advance.
By definition $$p_{-}=\sup\{a\,:\,\mu\{x\,:\, p(x)<a\}=0$$ If $p_{-}=\infty$ then for every $a$ there holds $\mu\{x\,:\, p(x)<a\}=0.$ We have $$\{x\,:\, p(x)<\infty\}=\bigcup_{n=1}^\infty \{x\,:\, p(x)<n\}\\ = \bigcup_{n=1}^\infty \{x\,:\, n-1\le p(x)<n\}$$ Thus $$\mu \{x\,:\, p(x)<\infty\}\\ = \sum_{n=1}^\infty \mu\{x\,:\, n-1\le p(x)<n\}\\ \le \sum_{n=1}^\infty \mu\{x\,:\, p(x)<n\}=0$$ Therefore $p(x)=\infty$ almost everywhere.