Let $(\Omega,\mathcal A,\mu)$ be a measure space and $(f_n)$ a sequence of $\mu$-integrable functions, which converges uniformly to $f:\Omega\to\mathbb R$.
1) If $\mu$ is finite, $f$ is also integrable.
2) If $\mu$ is $\sigma$-finite, $f$ needn't be integrable. In case $f$ is integrable, $\lim_{n\to\infty}\int f_n d\mu$ and $\int f d\mu$ needn't be the same.
I tried showing 1) like this:
Since $f_n\to f$ uniformly we have $\forall x\in\Omega \forall\varepsilon>0\exists N_0\in\mathbb N$ such that $\forall n\geq N_0$: $$|f_n-f|<\varepsilon$$
From this I got the following estimate:
$$\left|\int f_nd\mu -\int fd\mu\right|\leq \int |f_n-f|d\mu<\int \varepsilon d\mu\leq \varepsilon\mu(\Omega)$$
Since this holds for all $\varepsilon>0$ (and $\mu(\Omega)$ is finite) we deduce $\lim_{n\to\infty}\int f_nd\mu = \int fd\mu$.
To show that $f$ is indeed integrable I tried showing that $|f|$ is integrable:
$$\int |f|d\mu=\int |f-f_n+f_n|d\mu\leq \int |f-f_n|d\mu + \int |f_n|d\mu\leq \varepsilon\mu(\Omega)+c<\infty$$
Since $f_n$ is integrable, $|f_n|$ is integrable, too, and the integral is finite and hence the integral over $|f|$ is finite and hence $f$ is integrable.
Is this correct?
And how do I show 2)? I tried thinking of some counter-examples, but I couldn't think of any. Can anyone else help me out here?
Thanks!