If $f_n \rightarrow f$ uniformly and $f_n$ are measureable, integrable functions with $\mu$ $\sigma$-finite. Then $f$ has not to be integrable. Note that $f_n: \Omega \rightarrow \mathbb{R}$. I am new to measure-theory and got few problems with the idea and the right writing. I am using the definition $$\int_{\Omega} f d\mu=\sum_{n=1}^\infty a_n \mu(A_n).$$
My idea was to construct $f_n$ such that for $f$ we have $\Omega=\{x\in \mathbb{R}|x\geq 1\}$, $a_n=1/n$ and $A_n=[n,n+1]$ so that $\mu([n,n+1])=1$ and the integral would equal the harmonic series which diverges. How can I construct such $f_n$? And how can I determine those $a_n$ in general? thanks for your help.
With $f=\displaystyle\sum_{n}\dfrac{1}{n}\chi_{[n,n+1)}$, $f_{n}=\displaystyle\sum_{1\leq k\leq n}\dfrac{1}{k}\chi_{[k,k+1)}$, then $|f_{n}-f|\leq\dfrac{1}{n+1}$ and $\|f\|_{L^{1}([1,\infty))}=\displaystyle\sum_{n}\dfrac{1}{n}=\infty$.