# measurable spaces and $L^p$

We need to prove that $g\in L^p \; \forall\;\;p \; \text{s.t.} \; 0 <p <1$, but that $g$ need not be in $L^1$ given that following:

Suppose that $(Y, M, \eta)$ is a finite measure space, $g\in L^{+}(Y,M)$ and $\Lambda_{g}(v)\leq (v^{-1}) \quad \forall v>0$, where $\Lambda_{g} : (0, \infty) \rightarrow [0,\infty]$ by: $$\Lambda_{g}(v) =\eta(\{x:g(x)>v\}) \quad\text{and}\quad\int(g(x))^{p}d\eta(x) =p \,\int_{0}^{\infty} v^{p-1}\Lambda_{g}(v)dv$$

This is what I was trying to do but I don't know whether its correct. Suppose $p=1$, observe that the above integral equation becomes $\int_{0}^{\infty} \Lambda_{g}(v)dv$ , and then I used definition of $\Lambda_{g}$. Is is the right thing to do so far?

-
What have you tried so far? Also - many users here consider it rude to copy questions directly from the textbook/source, and much prefer if you remove imperative orders (such as "Show that..."). If you include your work, give some context, and rephrase, I bet that many more people will be willing to help you. But welcome to MSE! –  mixedmath Nov 13 '11 at 3:16
In addition to mixedmath's comment, a little "air" in your question wouldn't hurt either, everything seems glued together in the text –  Patrick Da Silva Nov 13 '11 at 3:28
This is what I did, observe that if we suppose p=1, the given integral equation becomes easy, so I want to prove using that. a contradiction. @ Patrick Da Silva, what do you think. –  Heavy Nov 13 '11 at 3:36

I hope this works. Firstly substitute $\Lambda _{g}$ into the given integral equation. Secondly, Use a theorem on Chebyshev's to find a bound on $\Lambda _{g}$. Thirdly, use DCT.