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I have been reviewing measure theory and am a bit confused over which integral definition to use:

If we let $(X, A, \mu)$ be a measure space then

Firstly we have an integral of a non-negative simple function: $\int_{X}f d\mu := \sum_{i=1}^{N}a_{i}\mu(A_{i})$ where $f:= \sum_{i=1}^{N}a_{i}\chi_{A_{i}}$

Then we progress to the integral of a non-negative measurable function. This is the sup over all non-negative simple functions.

Finally, we progress to an integral of a general measurable function which we split into the positive and negative part.

I understand all of the above.

When it comes to apply this integration theory to any problems I always get confused what definition to use. If we are given a non-negative measurable function or a general measurable function that we want to integrate, is the general approach to rewrite it, to be a simple function. Then simply apply that definition?

As a concrete example of what I mean I found the following exercise online

Let $(X, A, \mu)$ be a measure space and $f: X \rightarrow [0, \infty]$ be a measurable function such that $0 < c:= \int_{X}f d\mu < \infty$. Prove that

$$\lim_{n \to \infty} \int_{X} n \log \Big(1 + \frac{f^{\alpha}}{n^{\alpha}}\Big)d \mu = c $$ (if alpha = 1)

Now I know the different theorems, MCT, DCT, fatou etc. I understand the theory but I have no idea how to apply it to the above problem since the integral doesn't contain any characteristic functions and/or simple functions...

Please let me know if things aren't clear as I really want to be able to apply this theory to problems!

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  • $\begingroup$ the general definition of being integrable is that $f$ is measurable and $\int|f|$ is finite, you can achieve this conclusion using various theorems. For you case note that $n\log\left(1+\frac{f^\alpha }{n^\alpha }\right)$ is non-negative and increasing so you can use the monotone convergence theorem $\endgroup$
    – Masacroso
    Mar 14 '20 at 10:05
  • $\begingroup$ @Masacroso the non-negative is obvious. To show it is increasing can I just say that if we write $f_{n} = n\log\left(1+\frac{f^\alpha }{n^\alpha }\right)$ then we will have that $f_{n} \leq f_{n+1}$? Applying MCT lets us bring the limit inside the integral so I can rewrite it as $\int_{X} \lim_{n \to \infty} \log(1+\frac{f}{n})^{n} = \lim_{n \to \infty} \log(e^{f}) = f$ (assuming alpha = 1). If alpha isn't 1 then the limit becomes $\lim_{n \to \infty} \log(e^{\alpha f})$?? $\endgroup$
    – VBACODER
    Mar 14 '20 at 11:15
  • $\begingroup$ yes, you can do this. For $\alpha >1$ just write $n=n^{1-\alpha }n^{\alpha }$ and apply the dominated convergence theorem $\endgroup$
    – Masacroso
    Mar 14 '20 at 11:18
  • $\begingroup$ @Masacroso I will try and apply the DCT however i always get stuck on proving that a function meets the necessary conditions to use it. $\endgroup$
    – VBACODER
    Mar 14 '20 at 11:25
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Suppose that $\alpha >1$, then for some non-negative function $f$ we have that $$ n^{1-\alpha }\ln \left(1+\frac{f(x)^{\alpha }}{n^{\alpha }}\right)^{n^\alpha }\leqslant \ln \left(1+\frac{f(x)^{\alpha }}{n^{\alpha }}\right)^{n^\alpha }\leqslant f(x)^{\alpha } $$ for all $x$ and all $n\in \Bbb N\setminus\{0\} $, then you can apply the dominated convergence theorem whenever $\int f(x)^{\alpha }\mathop{}\!d x<\infty $. When $\alpha =1$ you can use the monotone convergence theorem to finish.

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  • $\begingroup$ sorry I'm confused. To apply DCT don't I need to show the following two conditions: 1): The sequence of functions converge pointwise a.e. and 2): There is some other integrable function $g$ such that for all $i \in \mathbb{N}$ $|f_{i}| \leq g$ a.e.? I also don't fully understand why the middle term in your inequality is to the power $n^{\alpha}$ and not just $n$? Apologies, measure theory really confuses me when it comes to applying it to actual problems. $\endgroup$
    – VBACODER
    Mar 14 '20 at 11:38
  • $\begingroup$ @VBACODER if $f(x)\neq\infty$ then the sequence for such $x$ of $$n^{1-\alpha }\ln \left(1+\frac{f(x)^{\alpha }}{n^{\alpha }}\right)^{n^\alpha }$$ converges to zero $\endgroup$
    – Masacroso
    Mar 14 '20 at 11:40
  • $\begingroup$ I think I need to review the basics as I'm not seeing it... My analysis knowledge is awful :( $\endgroup$
    – VBACODER
    Mar 14 '20 at 11:42
  • $\begingroup$ for 2): this is just because $n=n^{1-\alpha }n^{\alpha }$ $\endgroup$
    – Masacroso
    Mar 14 '20 at 11:42
  • $\begingroup$ This is where I get confused. Because we also need to show that $g$ is an integrable function. Also, how do we then know that $n=n^{1- \alpha}n^{alpha} \geq |f|$ a.e. [assuming you're talking about (2) that I said]. $\endgroup$
    – VBACODER
    Mar 14 '20 at 11:47

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