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!
Best Answer
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.