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Let $(X,\Omega,\mu)$ be a measure space and $M^+(X,\Omega)$ denote the set of all non-negative real valued measurable functions.

If $f \in M^+(X,\Omega)$ and $0< \int f \mathrm{d}\mu < \infty$ then $$\lim_{n\rightarrow\infty} \int n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)\mathrm{d}\mu = \infty.$$

The hint given says use Fatou's Lemma. I am not really too sure exactly what Fatou's Lemma will do to help me solve this:

If I let $$f_n = n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)$$ then $f_n \in M^+(X,\Omega)$ and so Fatou's Lemma says:

$$\int \liminf f_n \mathrm{d}\mu \leq \liminf \left( \int f_n \mathrm{d}\mu \right).$$

To be honest, $\liminf f_n$ looks daunting to me as I know it as $$\sup_{m \in \mathbb{N}}\left( \inf_{n \geq m} f_n \right)=\sup_{m \in \mathbb{N}}\left( \inf_{n \geq m}\left[ n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)\right] \right).$$

My actual attempt at solving the problem was looking at the set $E = \{ x \in X : 1+\left(\frac{f}{n}\right)^{\frac{1}{2}} \geq e \}$ and then breaking $$\int f_n \mathrm{d}\mu = \int_E f_n \mathrm{d}\mu + \int_{X\setminus E}f_n\mathrm{d}\mu$$ and then we have $$\int_E n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right) \mathrm{d}\mu \geq \int_{E} n \mathrm{d}\mu=n\mu(X)$$ and assuming we have a space with non-zero measure (Do some axioms include just looking at measures that give the space positive measure?) then we'd have $\int f_n \mathrm{d}\mu \geq n\mu(X)$ and by taking limits of both sides (and as long as $\mu(E) \neq 0$) we'd get $\lim \int f_n \mathrm{d}\mu = \infty$ as needed. Problem is I wasn't sure what to do in the case that $\mu(E)=0$.

I would like to add I am interested in exercises to help work on improving my comfort and understanding of $\limsup$ and $\liminf$ if possible.

Thanks very much!

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You can check that for $a>0$ the function $$ \varphi(t)=t\log\left(1+\left(\frac{a}{t}\right)^{1/2}\right) $$ is strictly increasing on $[1,+\infty)$ and tends to infinity. Unfortunately you had to dig up to second derivative to prove this. Thus for each $x\in X$ you have strictly increasing to infinity sequence $\{f_n(x):n\in\mathbb{N}\}$, so $$ \liminf\limits_{n\to\infty} f_n(x)=\lim\limits_{n\to\infty} f_n(x)=+\infty $$ The rest is clear.

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  • $\begingroup$ @Wishingwell You are welcome! $\endgroup$ – Norbert May 25 '13 at 2:40
  • $\begingroup$ Can I ask how you used the second derivative to show $\varphi'(t)>0$ on $[1,+\infty)$? It might be something I haven't seen for a while $\endgroup$ – Frudrururu May 25 '13 at 7:39
  • $\begingroup$ You can check that $\varphi''<0$ so $\varphi'$ is decreasing but $\lim\limits_{t\to+\infty}\varphi'(t)=0$. Hence $\varphi'>0$ Thus $\varphi$ is increasing. $\endgroup$ – Norbert May 25 '13 at 8:40
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If $\lim_{n \to \infty} f_n$ exists, then $\lim$ and $\liminf$ coincides. Therefore, by Fatou's lemma, $\lim_{n\to \infty}\int n \log(1+(f/n)^{1/2})d\mu \ge \int \lim_{n\to\infty}n\log(1+(f/n)^{1/2})d\mu = \int \lim_{n\to\infty}(n/f)^{1/2}\log(1+(f/n)^{1/2})n^{1/2}f^{1/2}d\mu$ Since $(n/f)^{1/2}\log(1+(f/n)^{1/2})$ goes to 1 as $n\to\infty$ whenever $f>0$, the integrand goes to infinity as $n\to\infty$ for $f>0$. Since the integral is positive, the measure of $\{x:f(x)>0\}$ is nonzero, so the resulting integral also goes to infinity.

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