Hi I am struggling with a question but really I am struggling more with the concepts behind it so any help would be appreciated.
Q/ Let $$f(x)=\begin{cases} x^{-\frac{1}{2}} & ,x\in(0,1) \\ 0 & , \text{otherwise} \end{cases}$$ Let $r_k$, k={1,2,3...} be an enumeration of all rationals and set $$ g(x)=\sum_{k=1}^{\infty}2^{-k}f(x-r_k) $$ Prove g is lebesgue integrable.
So if we say $\lambda$ is the lebesgue measure, the question is asking us to show that $\int g \;d\lambda$ is defined. It is clear g is not a simple function so I am thinking the first step is to show that g is a measurable function but then the question seems ambiguous, what sigma algebra do I use on $\mathbb{R}$? Is the standard to use the Borel sets? or maybe I should use the set of all Lebesgue measurable functions?
Then since g is non negative the definition of the integral says
$\int g \;d\lambda=sup\{\int f \;d\lambda\; :0\leq f\leq g\}$ where f is simple and measurable, so then I show this supremum exists and I am done?
Thanks
Since you're struggling with the concepts, first you have to show that each function $g_n(x)=2^{-n}f(x-r_n)$ is ($\mathcal{L},\mathcal{B}_{\mathbb{R}}$) measurable (the inverse image of a Borel set is in the Lebesgue $\sigma$-algebra). It suffices to check this for a set $(a,\infty), a\in\mathbb{R}$.
From there, you need to apply a limit theorem to the sum $s_n(x)=\sum_1^n g_n(x)$ (if two functions are measurable, their finite sum is measurable) that says that the limit of measurable functions is again measurable.
As a hint for the next part, verifying directly from the definition of the integral is difficult. I would say compute the integral $\int_{\mathbb{R}} g_n d\lambda$ first (just using basic calculus), and then think about applying one of the convergence theorems for Lebesgue integrals.