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Let $f:X\rightarrow [0,\infty[$ be a measurable function that is greater than or equal to $1$ for every $x\in X$ and $\mu$ be a positive measure on $X$. Consider the function $g:]0,\infty[\rightarrow [0,\infty]$ that sends $p$ to $\int_X f^pd\mu$, must $f$ be continuous ?

I think the answer is yes, but I did not succeed in finding anything useful. I prefer hints rather than full answers

Thanks in Advance

Amr
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  • Please do not close this as a duplicate of http://math.stackexchange.com/questions/133773/is-p-mapsto-f-p-continuous. They are not duplicates – Amr Dec 03 '13 at 20:05
  • Could you clarify some things? Do you mean is f continuous in x for each p? What does g, a mapping of p, have to do with this? And do we know if f or $f^p$ is integrable on $[0, \infty]$? – Betty Mock Dec 03 '13 at 20:16
  • @BettyMock No. We do not know that $f^p$ is continuous. What do you mean by f continuous in x for each p ? – Amr Dec 03 '13 at 20:18
  • I don't know what I meant either. Dan Fisher has dealt with it, and I now know what you were asking. – Betty Mock Dec 04 '13 at 05:24

1 Answers1

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No, $g$ need not be continuous. It can have a jump discontinuity if you have an $f$ such that $\int_X f^{p_0}\,d\mu < \infty$, but $\int_X f^p\,d\mu = \infty$ for all $p > p_0$. An example for such an occurrence is

$$f(x) = \frac{1}{x(\log x)^2}$$

on $(0,1/4)$. $f \in L^1((0,1/4))$, but $f \notin L^p((0,1/4))$ for all $p > 1$.

However, that is the only type of discontinuity that can occur, for points where $g$ is finite in a neighbourhood of $p$, the monotone convergence theorem asserts the continuity.

Daniel Fischer
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  • Hi Daniel, as usual you offer a lot of help to me. Do you agree with the answer of Zarrax in this post: http://math.stackexchange.com/questions/133773/is-p-mapsto-f-p-continuous – Amr Dec 03 '13 at 20:16
  • Yes, that's right. The important difference is that there only the interval where $\int f^p < \infty$ was considered. On that interval we have continuity, and we also have continuity on the interval where $\int f^p = \infty$. But where the two intervals meet, we can have a jump. – Daniel Fischer Dec 03 '13 at 20:21
  • Aaaah. I misread the question and didn't see the condition $|f|_p < \infty$ for all $1\leq p < p'$. Now it makes sense. Thanks Daniel +1 – Amr Dec 03 '13 at 20:25