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Let $(X, \mathcal{A}, \,u)$ be a measurable space and $f: X \rightarrow [0, +\infty]$ a measurable function such that $\int_Xf \mu =1$. Show that $$\lim_{n\to\infty} \int_X n\ln\left(1 + \frac{f}{n}\right)\, d\mu = 1$$

I am new to measure theory, so I still lack any intuition in it. I managed to bound this integral by 1. For a fixed $n \in \mathbb{N}^*$, we have: $$ \int_X nln(1 + \frac{f}{n})d\mu \leq \int_X n( \frac{f}{n})d\mu = \int_xfd\mu =1$$

Now, this is not sufficient, and I wonder two things: Can I say, that for an $n \in \mathbb{N}$ big enough, we have $|\frac{f}{n}|<1$ even though $f$ could take the value of $+\infty$? And secondly, if the answer to the first question is yes, could I then use Taylor series? This way I could appriximate $\ln(1 + \frac{f}{n})$ to $\frac{f}{n} + o(\frac{f^2}{n^2})$.

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    $\begingroup$ Taylor series are a good idea in practice; if I were given that expression in "real life" (i.e., without the solution), I would develop in Taylor series to get an idea of what the result might be. Unfortunately, the difficulties you point out (i.e. the series for $\log(1+x)$ only converges for $|x|<1$) are real and will need to be addressed if you want to use Taylor series to rigorously prove the result. One possibility is to approximate $f$ with a truncated version of it: [...] $\endgroup$ Commented Feb 7, 2018 at 17:10
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    $\begingroup$ $$f_{m, n}(\omega):=\begin{cases} f(\omega), & f(\omega)<n \\ 1, & f(\omega)\ge 1.\end{cases}$$ Here, however, this method is complicated as you end up computing a limit over two indices $m, n$. $\endgroup$ Commented Feb 7, 2018 at 17:15

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We have $\log(1+u)\leq u$ for $u\geq 0$, so $\log(1+f/n)\leq f/n$, so $n\log(1+f/n)\leq f$. Now $n\log(1+f/n)\rightarrow f$ a.e. as $n\rightarrow\infty$, so by Lebesgue Dominated Convergence Theorem we have $\lim_{n\rightarrow\infty}\displaystyle\int_{X}n\log(1+f/n)d\mu=\displaystyle\int_{X}fd\mu=1$.

Note that $\displaystyle\int_{X}fd\mu=1<\infty$, then $f<\infty$ a.e.

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