Suppose $\mu$ is a positive masure on $X$, $f:X\to [0,\infty]$ is measurable, $\int \limits_{X}fd\mu=c$, where $0<c<\infty,$ and $\alpha$ is a constant. Prove that $$\lim \limits_{n\to \infty}\int \limits_{X}n\log [1+(f/n)^{\alpha}]d\mu= \begin{cases} \infty, & \text{if} \quad0<\alpha<1, \\ c, & \text{if} \quad\alpha=1, \\ 0, & \text{if} \quad 1<\alpha<\infty. \end{cases}$$

Remark: $[\cdot]$ is not integer part!

Proof: $\color{blue}{Case \quad \alpha=1}$. Consider functions $f_n(x):X\to [0,\infty]$ defined by $f_n(x)=n\log \left[1+\dfrac{f(x)}{n}\right]$. It's easy to check that $0\leqslant f_1\leqslant f_2\leqslant \dots \leqslant f$ on $X-S$ where $S=\{x\in X:f(x)=\infty\}$ and note that $\mu(S)=0$ (otherwise $\int \limits_{X}fd\mu=\infty$ which is contradiction). Also $f_n$ is measurable for each $n$ since it's a compostion of continuous and measurable functions. Using Monotone Convergence Theorem we get: $$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}\lim \limits_{n\to \infty}f_nd\mu=\int \limits_{X}fd\mu=c.$$

$\color{blue}{Case \quad 0<\alpha<1}$. Using Fatou's lemma to functions $f_n=n\log[1+(f/n)^{\alpha}]$ which is measurable in $X-S$ for each $n$ we get the following inequality:

$$\liminf \limits_{n\to \infty} \int \limits_{X}f_nd\mu\geqslant \int \limits_{X}\liminf \limits_{n\to \infty} f_nd\mu$$

Since $\int \limits_{X}fd \mu=\int \limits_{X\setminus S}fd \mu=c$ then the set $E=\{x\in X-S: f(x)>0\}$ has positive measure. And $$\int \limits_{X}\liminf \limits_{n\to \infty} f_nd\mu=\int \limits_{X\setminus S}\liminf \limits_{n\to \infty} f_nd\mu=\int \limits_{E}+\int \limits_{(X\setminus S)\setminus E}=$$ Since on $(X\setminus S)\setminus E$ we have that $f_n(x)=0$ $$=\int \limits_{E}\liminf \limits_{n\to \infty} f_nd\mu=+\infty$$ since $\liminf \limits_{n\to \infty} f_n=+\infty$ on $E$ and $\mu(E)>0.$

$\color{blue}{Case \quad \alpha>1}$. Our functions $f_n$ are measurable and non-negative on $X-S$ and $f_n(x)\to 0$ as $n\to \infty$ on $X-S$. Using derivative test we can show that $f_n\leqslant \alpha f$ for $n\in \mathbb{N}$ on $X-S$. Note that $\alpha f\in L^1(\mu)$. By Dominated Convergence theorem $$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}\lim \limits_{n\to \infty}f_nd\mu=0.$$ Here we use that $\mu(S)=0$ because the measure of zero set in negligible in integration.

Sorry if this topic is repeated but I would be thankful if anyone checks out my solution.

  • $\begingroup$ Your proof looks good. I don't know what the derivative test is in the $\alpha>1$ case, but it is easy to establish by splitting into ${x \over n} \le 1$ and ${x \over n} > 1$ cases. $\endgroup$
    – copper.hat
    Jun 12, 2016 at 18:47
  • $\begingroup$ @copper.hat, I do not quite understand you. Can you demonstrate your approach? It would be great for me to know the new way of solution of this $\endgroup$
    – RFZ
    Jun 13, 2016 at 3:48
  • $\begingroup$ What is the 'derivative test'? If ${x \over n} \ge 1$, then $f_n(x) \le n \log ((1+{x \over n})^\alpha) = \alpha n \log (1+{x \over n}) \le \alpha x$, and if ${x \over n} < 1$ then $f_n(x) \le n \log (1+{x \over n}) = x$ and so $f_n(x) \le \alpha f(x)$. $\endgroup$
    – copper.hat
    Jun 13, 2016 at 5:29
  • $\begingroup$ @copper.hat, But you didn't show that $f_n(x)\leqslant \alpha f$ in the first case?! Where did you use that $x/n\geqslant 1$? $\endgroup$
    – RFZ
    Jun 13, 2016 at 10:33
  • $\begingroup$ @copper.hat, are you sure that you had not any typo? $\endgroup$
    – RFZ
    Jun 13, 2016 at 10:37

1 Answer 1


This is not an answer, just an elaboration of a comment:

Suppose $\alpha>1$.

(i) If ${f(x) \over n} \ge 1$ we have $f_n(x) \le n \log ((1+{f(x) \over n})^\alpha) = \alpha n \log (1+{f(x) \over n}) \le \alpha f(x) $.

(ii) If ${f(x) \over n} < 1$ we have $f_n(x) \le n \log (1+{f(x) \over n}) = f(x)$.

Hence $f_n(x) \le \alpha f(x)$.


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