# Prove the following statements are equivalent

Prove the following statements are equivalent:

1. for every $$f\geq 0$$,$$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$$
2. $$\mu$$ is finite measure and X is not the union of infinitely many sets of positive measure.

I've tried to assume that (2) doesn't holds and I managed to prove that $$\int_{X}fd\mu$$ doesn't converge but I didn't managed to prove for the other integral.
any for the second part I had nothing

• The difficult direction is $1 \implies 2$ (or "not 2" $\implies$ "not 1", as you attempted). However, you should be able to prove that $2 \implies 1$ without too much trouble. Dec 10, 2020 at 16:21
• Keep in mind that for the "not 2" $\implies$ "not 1", you should attempt to prove that if 2 fails to hold, then there exists a function $f\geq 0$ for which one integral converges but the other does not. It does not sounds like this is what you were trying to do Dec 10, 2020 at 16:23
• @BenGrossmann I got the telepathic powers on this one: $X$ is not the union of infinitely many sets of positive measure. Dec 10, 2020 at 17:10
• To my own question: discrete measures Dec 10, 2020 at 17:32
• In fact, $(2)$ entails a stronger result than $(1)$. Let assume $(2)$. Given any $f \geq 0$ and given any $n \in \mathbb{N}$ let $$A_n=\{x \in X: n \leq f(x) < n+1 \}$$ Clearly $\{A_n\}_n$ is an infinite family of disjoint sets and $X=\bigcup_nA_n$. Since $X$ is not the union of infinitely many sets with positive measure, it is easy to prove that there is $n_0 \in \mathbb{N}$ such that if $n \geq n_0$, $\mu(A_n)=0$. So $f<n_0$ $a.e.$. So $e^f<e^{n_0}$ $a.e.$. So we have $\int_X f d\mu < n_0\mu(X)<\infty$ and $\int_X e^f d\mu < e^{n_0}\mu(X)<\infty$. Dec 10, 2020 at 21:01

Let us prove the result.

Let $$(X,\Sigma, \mu)$$ be a measure space.

$$(1 \Rightarrow 2)$$. Suppose for every $$f\geq 0$$,$$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$$.

Take $$f=0$$. Clearly, $$\int_{X}fd\mu=0<\infty$$. So $$\mu(X)= \int_X 1 d\mu= \int_X e^0 d\mu= \int_X e^f d\mu < \infty$$

Now, let us prove that $$X$$ is not the union of infinitely many sets of positive measure.

Suppose that $$X$$ is the union of infinitely many sets of positive measure. Since $$\mu(X)<+\infty$$, it follow that for all $$\delta >0$$ there is $$E \in \Sigma$$ such that $$\mu(E)<\delta$$. It follows, from lemma $$1$$ and lemma $$2$$ in the Remark 1, that $$L^2(X) \subsetneq L^1(X)$$. Take a function $$f \in L^1(X)$$ such that $$f \notin L^2(X)$$. But then we have that $$|f|\geq 0$$, $$\int_{X}|f|d\mu<\infty$$ and $$\int_X e^{|f|} d\mu > \int_X \left (1+|f| + \frac{1}{2}|f|^2 \right) d\mu \geqslant \frac{1}{2} \int_X |f|^2 d\mu =+\infty$$ Contradiction to $$(1)$$. So $$X$$ is not the union of infinitely many sets of positive measure.

$$(2 \Rightarrow 1)$$ In fact, $$(2)$$ entails a stronger result than $$(1)$$. Let us assume $$(2)$$. Given any $$f \geq 0$$ and given any $$n \in \mathbb{N}$$ let $$A_n=\{x \in X: n \leq f(x) < n+1 \}$$ Clearly $$\{A_n\}_n$$ is an infinite family of disjoint sets and $$X=\bigcup_nA_n$$. Since $$X$$ is not the union of infinitely many sets with positive measure, it is easy to prove that there is $$n_0 \in \mathbb{N}$$ such that if $$n \geq n_0$$, $$\mu(A_n)=0$$. So $$f $$a.e.$$. So $$e^f $$a.e.$$. So we have $$\int_X f d\mu < n_0\mu(X)<\infty$$ and $$\int_X e^f d\mu < e^{n_0}\mu(X)<\infty$$. Then $$(1)$$ follows immediately.

Remark 1:

Lemma 1. Let $$(X,\Sigma, \mu)$$ be a measure space. If $$\mu(X)<+\infty$$ is finite then $$L^2(X) \subseteq L^1(X)$$.

Proof Since $$\mu(X)<+\infty$$, we have $$\chi_X \in L^2(X)$$. Then for any $$f\in L^2(X)$$, applying Hölder inequality, we have $$\|f\|_1= \|f\, \chi_X \|_1\leqslant \|\chi_X\|_2 \|f\|_2 =(\mu(X))^{1/2} \|f\|_2$$

Lemma 2. Let $$(X,\Sigma, \mu)$$ be a measure space. If $$L^1(X) \subseteq L^2(X)$$, then there is $$\delta >0$$ such that, for any $$E\in \Sigma$$, if $$\mu(E)>0$$ then $$\mu(E) > \delta$$.

Proof: Let $$T: L^1(X) \rightarrow L^2(X)$$ be the inclusion map, that is $$T(f)=f$$. $$T$$ is obvious linear. Using the Closed Graph Theorem, we can prove $$T$$ is continuous (bounded).

In fact, let $$\{f_n\}$$ be a sequence in $$L^1(X)$$ which converges to $$f$$ in the $$L^1$$ norm, and to $$g$$ for the $$L^2$$ norm. We extract a subsequence $$\{f_{n_j}\}$$ which converges to $$f$$ $$a.e.$$; this subsequence still converges to $$g$$ for the $$L^2$$ norm; now extract from this subsequence a new subsequence which converges to $$g$$ almost $$a.e.$$. So we have a subsequence which converges $$a.e.$$ to $$f$$ and $$g$$, hence $$f=g$$, and, since both $$L^1(X)$$ and $$L^2(X)$$ are Banach spaces, by the Closed Graph Theorem we conclude that $$T$$ is continuous (bounded).

Since $$T \neq 0$$, we have that $$0<\|T\| <+\infty$$.

Now, let us prove that, given any $$E \in \Sigma$$ such that $$\mu(E)>0$$, $$\mu(E) \geqslant \frac{1}{\|T\|}$$.

If $$\mu(E)=+\infty$$, then it trivial that $$\mu(E) \geqslant \frac{1}{\|T\|}$$.

Suppose $$0<\mu(E)<+\infty$$, let $$f_E= \frac{1}{\mu(E)} \chi_E$$ It follows immediately that $$\|f_E\|_1=1$$ and $$\|f_E\|_2=\frac{1}{\mu(E)}$$. So $$\frac{1}{\mu(E)}=\|f_E\|_2 =\|T(f_E)\|_2 \leqslant \|T\| \|f_E\|_1 =\|T\|$$ So $$\mu(E) \geqslant \frac{1}{\|T\|}$$ Taking $$\delta = \frac{1}{2\|T\|}$$, we have $$\mu(E) > \delta$$.

Remark 2: In fact, we proved a slightly more general result. We proved that he following statements are equivalent:

1. for every $$f\geq 0$$,$$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$$
2. $$\mu$$ is finite measure and X is not the union of infinitely many sets of positive measure.
3. for every $$f\geq 0$$,$$\int_{X}fd\mu<\infty$$ and $$\int_{X}e^{f}d\mu<\infty$$