I am stuck with Rudin's Real and Complex Analysis, Chapter 3 problem 4 question 3. It reads:
Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and $\phi(p)=\int_{X}|f|^{p}d\mu=|f|^{p}_{p}$ Let $E=\{p:\phi(p)<\infty\}$. Assume $|f|_{\infty}>0,0 . Q: By previous question we know $E$ is connected. Is $E$ necessarily open? Closed? Can $E$ consist of a single point? Can $E$ be any connected subset of $(0,\infty)$? I am really stuck. I want to show $E$ is open and can be any open subset of $X$, but I do not know how to prove it. Suppose $\phi(a)<\infty$, I do not know how to prove $\phi(a\pm \epsilon)<\infty$ as well when $|f|_{\infty}=\infty$ (otherwise I can use Holder's theorem to bridge) update: Someone asked me why $E$ is connected. here is the proof: Q:If $r , $r\in E, s\in E$, then $p\in E$. Let $X_{1}$ be the part $f$'s absolute value is greater than $1$, and $X_{2}$ be the part $f$'s absolute value is less to $1$. Let $X_{3}$ be the part where $f$'s absolute value equal to 1. Then we have $ \int_{X}|f|^{p}=\int_{X_{1}}|f|^{p}+\int_{X_{2}}|f|^{p}+\int_{X_{3}}|f|^{p} $ The first part is less than the part in integrating $|f|^{s}$, the third part is less than that of $|f|^{r}$, and the middle part is not changing at all. Thoughts: One way of attacking this is by Holder's inequality. We have $\int_{X}f^{a+\epsilon}d\mu<\int_{X}f^{a}d\mu *|f^{\epsilon}|_{\infty}$ And so if I can bound $|f^{\epsilon}|_{\infty}$ then I will be done. However, $|f|_{\infty}=\infty$ is entirely possible (for example $\frac{1}{x^{2}}$ on $\mathbb{R}$), and this proof cannot go through. Another way is to decompose $X$ into 3 parts as before. Then we only need to prove the statement makes sense for $|f|>1$ and $|f|<1$ respectively. Then further we only need to prove $+\epsilon$ on one direction and $-\epsilon$ on the other direction. In the former case $X_{1}$ must have finite measure, and the proof would work if $|f|_{\infty}<\infty$. But I have no idea how to prove the second case.