$μ$ is $σ$-finite iff $L^p(X)$ contains a strictly positive function.

Question

Let $(X,M, μ)$ be a measure space and $0 < p < ∞$. Prove that, $μ$ is $σ$-finite iff $L^p(X)$ contains a strictly positive function.

My Work:

If I suppose $L^p(X)$ contains a strictly positive function $f$, then $f^p>0$ on $X$ and $\int_X f^p d\mu<\infty$. Then by a Proposition, $\{x:f^p(x)>0\}$ is $\sigma$- finite. That is $\mu(X)<\infty$. Hence, $μ$ is $σ$-finite. But I was stuck in proving the other direction. Can anybody please give me a hint?

Answer 1

Let $(X,M,\mu)$ be a measure space and $0<p<\infty$

1) If $\mu$ is σ-finite we have $$X=\bigcup_{n=1}^\infty A_n, \ \ \mu(A_n)<\infty .$$ Let us be define $$f(x)=\sum \limits_{n=1}^{\infty}\frac{1}{2^{\frac{n}{p}}}\frac{1}{(\mu(A_n)+1)^{\frac{1}{p}}}X_{A_n}(x),$$ where $X_{A_n}(x)$ is the characteristic function of $A_n$. We note $f(x)>0$ moreover $$\int \limits_{X}f^{p}(x)d\mu\leq\sum \limits_{n=1}^{\infty}\frac{1}{2^{n}}\frac{\mu(A_n)}{(\mu(A_n)+1)}\leq\sum \limits_{n=1}^{\infty}\frac{1}{2^{n}}<\infty,$$ then $f\in \mathbf L^{p}$.

2) If $\mathbf L^{p}$ contains a strictly positive function $f$. For each $n\in \mathbb N$ we consider the measurable set $$A_{n}=\{ x\in X: f^{p}(x)>\frac{1}{n}\}.$$ We observe:

i) $X=\bigcup_{n=1}^\infty A_n, \ \ \mu(A_n)<\infty .$

ii) $\mu(A_n)\leq n\int \limits_{A_n}f^{p}(x)d\mu \leq n\int \limits_{X}f^{p}(x)d\mu<\infty.$\ From i) and ii) $\mu$ is $\sigma$ finite.

Regards!!

Answer 2

Hint: Suppose $X$ is $\sigma$-finite. Then $X = \cup_{i=1}^\infty K_i$ such that $\mu(K_i) < \infty$ for all $i = 1, 2, \dots$. Let $\{a_i\}_{i=1}^\infty$ be a sequence of strictly positive real numbers, and define $f = \lim_{n\rightarrow\infty}\sum_{i=1}^n a_i 1_{K_i}(x)$, where $1_A(x)$ is the indicator function of the set $A$. What is $\int f^p d\mu$ in terms of the $a_i$ and $K_i$? Now can you choose $a_i$ to make $\int f^p d\mu < \infty$?