# real analysis – Interval of \$p\$ that makes the \$p\$ power of a function integrable

For a measurable real-valued function $$f$$ on $$(0,infty)$$ let $$P(f):={pin(0,infty):|f|^pinmathcal{L}^1(mathbb{R},mathcal{B},lambda)}$$, where $$mathcal{B}$$ is the Borel $$sigma$$-algebra. Show that for every subinterval $$J$$ of $$(0,infty)$$, which may be open or closed on the left and on the right, there is some $$f$$ such that $$P(f)=J$$. Hint: Consider functions such as $$x^a|log x|^b$$ on $$(0,1)$$ and on $$(1,infty)$$, where $$b=0$$ unless $$a=-1$$.

I don’t understand the hint. $$f(x)=x^anotinmathcal{L}^1(mathbb{R},mathcal{B},lambda)$$ for any $$a$$. The log part is also not integrable:

$$begin{equation} begin{split} & int_0^infty x^{-1}|log x|^bdx \ = & int_0^infty |log x|^bdlog x \ = & int_0^1 (-log x)^bdlog x+int_1^infty (log x)^bdlog x \ = & int_{-infty}^0 (-x)^bdx+int_0^infty x^bdx \ = & -int_{-infty}^0 (-x)^bd(-x)+int_0^infty x^bdx \ = & -int_infty^0 x^bdx+int_0^infty x^bdx \ = & int_0^infty x^bdx+int_0^infty x^bdx \ end{split} end{equation}$$

But for $$J=(0,infty)$$, I can find such a $$f(x)$$:
$$begin{equation} f(x) = begin{cases} x-1 &text{1

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