Let $X=\{x_1,x_2,\ldots\}$, and for any $A\subseteq X$ let $$\mu(A)=\sum_{x_n\in A}2^{-n}$$ Suppose $f$ is given by $f(x_n)=1/n!$. Find $$\int_X fd\mu$$
I tried to estimate from below by simple functions $s_k$ where $s_k(x_n)=1/n!$ for $n=1,2,\ldots,k$ and $s_k(x_n)=0$ otherwise. Then $\int_Xs_kd\mu=\sum_{i=1}^k2^{-i}/i!$.
I'm quite sure that $\int_Xfd\mu=\sum_{i=1}^\infty 2^{-i}/i!$. But since there could be other simple functions that are $\leq f$, how can I prove it rigorously?
Your sequence $\{s_k\}$ converges to $f$ monotonically. The monotone convergence theorem gives you the value of the integral.