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?



Your proof that $f$ is integrable by the approximation $\int|f| \leq \int|f-f_n| + \int|f_n|$ looks good, though on the previous estimate, you should start with $\int|f-f_n|$ without the previous term $\left| \int f - \int f_n\right|$ since before proving integrability of $f$, it doesn't make sense to write $\int f$ (and that term is unneeded for your conclusion).

For a counter-example, consider $f(x) = 1/x$ on $[0, \infty)$ and let $$f_n(x) = \begin{cases} f(x) & 0 \leq x \leq n \\ 0 & x > n \end{cases}$$

Then $\sup_{x \in [0,\infty)} |f(x)-f_n(x)| = \frac{1}{n} \to 0$, but $f$ isn't integrable (using Lebesgue measure and the standard Borel $\sigma$-algebra on $[0,\infty)$ ).


Let $\phi_n = \frac{1}{n} 1_{[n,2n)}$. Note that $\int \phi_n = 1$.

Let $f_n = \sum_{k=1}^n \phi_{2^k}$, and $f = \sum_{k=1}^\infty \phi_{2^k}$.

Then $\int f_n = n$, $f_n \to f$ uniformly, but $\int f = \infty$.


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