# Calculate an integral in a measurable space

Let $(X,\mathcal{M})$ a measurable set with measure $\mu$. Let $f$ be an integrable non negative function, such that $K:=\int_{E}f \mathrm d\mu<\infty$, where $E\in(X,\mathcal M)$. Let $\alpha>0$. Calculate the following integral

$\displaystyle \lim_{n\rightarrow\infty}\int_{E} n \ln\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)\mathrm{d}\mu$

I prove that for $\alpha=1$ the previous integral is $K$, and $\forall \alpha>1$ is zero. It follows by the identity $(1+x^{\alpha})\leq\alpha x$, $\forall x\geq0$ $\forall \alpha\geq1$. And the Dominated Convergence Theorem. But I don´t know how to take the case $0<\alpha<1$. Thanks in advance!

Suppose that $\alpha>1$. We take the sequece $\{f_{n}\}_{n\in\mathbb N}$ of measurable functions, $f_{n}:=n\ln(1+(f/n)^{\alpha})$. Moreover by the identity, for $\alpha>1$, we have the following $|f_{n}|=\left|n\log\left( 1+\left(\frac{f}{n}\right)^{\alpha}\right) \right|\leq n \alpha\frac{f}{n}=\alpha f$, $\alpha f$ is measurable. We need only to prove that$\{f_{n}\}_{n\in\mathbb N}$ converges pointwise to a function.

In this case $f\equiv 0$ $\forall x\in Dom(f)$, $\displaystyle\lim_{n\rightarrow \infty}n\log\left(\left(\frac{f}{n}\right)^{\alpha}+1\right)= \lim_{t\rightarrow 0}\frac{\log((ft)^{\alpha}+1)}{t}$. Making the change of variables $t=\frac{1}{n}$. And by L'Hopital, it arises $\lim_{t\rightarrow 0}\alpha f^{\alpha} \frac{t^{\alpha-1}}{(ft)^{\alpha}+1}=0$. ($\alpha-1>0$). Finally, using the Dominated Convergence Theorem, it leads to $\displaystyle\lim_{n\rightarrow \infty} \int f_{n} \mathrm d \mu=\int 0 \mathrm d\mu =0$

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Could you post your proof that it is zero for any $\alpha>1$? I suspect that rearranging it one should be able to prove that the integral actually diverges for any $0<\alpha<1$. – Giovanni De Gaetano May 13 '13 at 8:47

By a familiar property of $e$ and the exponential function, we have: $$\lim_{n\to\infty} n^\alpha \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) = f^\alpha$$

Hence: $$\lim_{n\to\infty} n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) = f^\alpha \lim_{n\to\infty} n^{1-\alpha}$$

When $0 < \alpha < 1$ and $f \ne 0$, the the limit diverges to $\infty$.

Apply Fatou's lemma: $$\int \liminf_{n\to\infty} n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) \,d\mu \le \liminf_{n\to\infty} \int n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) \,d\mu$$

Assuming $f$ is not zero a.e., the limit on the LHS diverges to $\infty$ for a set of positive measure. It follows that the limit on the RHS is $\infty$.

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Thanks!! Ayman Hourieh – Aleph May 13 '13 at 9:02
@Aleph You're welcome! – Ayman Hourieh May 13 '13 at 9:02