I try to proof a claim. It should be done first for indicator functions, then for step functions and finally for limits of increasing sequences of step functions. I'm not sure if I'm doing it right. So I hope somebody can have a quick look at it.
This is the claim that I wanna prove:
Claim: Let the function $g:[0,\infty) \rightarrow [0,\infty) $ be non-increasing and $\int_{0}^{\infty} g(t)dt<\infty$. Then $$\int_{0}^{t} g(t-x)dm(x) \rightarrow \frac{1}{\mu} \int_{0}^{\infty} g(x)dx \text{ as } t\rightarrow \infty.$$
i) I choose a set $A$ in $[0,\infty)$. So I set $A=[a,b]$, with $a \geq 0$ and $b< \infty$. Therefore my function $g$ is: $$g=1_A.$$ Since the function should be non-increasing, the interval $A$ should begin at zero. So $A=[0,b]$. Is that right? If yes, then I would be finished for indicator function, since I can apply a theorem (I could add it if somebody wants). For me it would be a big release if I would know if I choosed the set $A$ right!
ii) $g$ is now a step function. So it is of the form: $$g(x)=\sum_{i=0}^{n} 1_{A_i}(x) a_i$$, where $A_i$ are intervalls of $[0,\infty)$ and $a_i$ is a non-negative real number. But since $g$ is again non-increasing, the intervalls can only be of the form $A_i=[0,b_i]$. Is this again right? If yes, then I get by using linearity of the integral and i) the right result.
iii) Now suppose $g$ is an (arbitrary) function satisfying the condition of the claim. So my increasing sequence of step functions $f_n$ is of the form: $$f_n(x)= \begin{cases} g(x) , &x \leq n \\ 0, &x>n \end{cases}$$ So we have: $$ \lim_{n \rightarrow \infty } f_n (x) =g(x) \text{ and } f_n (x) \leq f_{n+1}(x) \leq g(x).$$ Therefore we can add the lemma about monotone convergence: $$\int \lim_{n \rightarrow \infty } f_n (x) = \lim_{n \rightarrow \infty } \int f_n (x).$$ So we get: $$\int_{0}^{t} g(t-x) dm(x) = \int_{0}^{t} \lim_{n \rightarrow \infty } f_n (t-x) dm(x)= \lim_{n \rightarrow \infty } \underbrace{ \int_{0}^{t}f_n (t-x) dm(x)}_{\rightarrow \frac{1}{\mu} \int_{0}^{\infty} f_n (x) dx}= \lim_{n \rightarrow \infty } \frac{1}{\mu} \int_{0}^{\infty} f_n (x) dx = \frac{1}{\mu} \int_{0}^{\infty} \underbrace{ \lim_{n \rightarrow \infty } f_n (x)}_{=g(x)} dx = \frac{1}{\mu}\int_{0}^{\infty} g(x) dx$$ if $t$ goes to infinity.
Is this also right?
I really hope that anybody can check if it is right or not what I did so far! I appreciate any advice, help or idea! Thank you so far!
You should clarify what $\mu$, $m$ are (in your comment you are nor really precise, I guess $X$ is not just some random Variable, but is related to the integral you're computing). Since your question is not really about that, I will answer the other parts.
First, you need to understand that you are not "choosing" $A$. $A \subset (0,\infty)$ should be an arbitrary set, such that $1_A$ is non-increasing. This means that $1_A(x) \geq 1_A(y)$ whenever $0 \leq x \leq y$. In other words $0 \leq x \leq y$ and $y \in A$ implies $x \in A$, which means $A$ is of the form $[0,a]$ or $[0,a)$ for some $a \geq 0$.
For the second part, the argumentation should be similarly. By the same argument the support of $g$ will be of the form $[0,a]$ or $[0,a)$. Assume that the $a_i$ are sorted increasingly. Then take $B_1:=[0,a]$ (or $[0,a)$) and $b_1=a_1$. Then consider $g-a_1 \cdot 1_{B_1}$ and do the same to get $B_2=[0,a']$ (or $[0,a')$) and take $b_2=a_2-a_1$. Going on like that you receive $B_i$ and $b_i$, such that $g=\sum_{i=1}^n b_i 1_{B_i}$. But now the functions $1_{B_i}$ are suitable to apply (i).
For (iii) your idea is good, but the functions $f_n$ aren't step functions in general. It is not too hard to find such functions though. There's a well known way to find a monotone sequence of simple functions $f_n$ converging pointwise to $g$ whenever $g$ is measurable by separating the y axis (you probably did this in class). You just need to verify why these functions are non-increasing step functions in your case.