Counter example for $\int_\mathbb{R} f_nd\mu\rightarrow\int_\mathbb{R} fd\mu$

Question

Assume $\{f_n\}$ and $f$ are Lebesgue measurable. Let $f_n$ be integrable $\forall n$, and assume $f_n\rightarrow f$ uniformly on $\mathbb{R}$. Assume $f$ is integral. Is it true that $\int_\mathbb{R} f_nd\mu\rightarrow\int_\mathbb{R} fd\mu$?

*I claim no (because I have problem with proving $\epsilon\times\mu(\mathbb{R})$ being finite if the above statement is true.) However, I can't come up with a counter example...

Answer 1

Consider the "melting ice cube" example (I like to have little mnemonics for remembering useful counterexmples): let $f_n := \frac{1}{n} \chi_{[0,n]}$. Then $f_n$ converges uniformly to zero, but $\int f_n = 1$ for all $n$, and so $$ \lim_{n\to \infty} \int f_n = 1 \ne 0 = \int \lim_{n\to\infty} f_n. $$ Note that if $\mu(X)$ is finite, then uniform convergence is good enough to get convergence in $L^1$ (which is, more or less, what you are trying to do).

Answer 2

Have $f_n$ equal to $\frac{1}{n}$ over $[a_n,a_n+n]$. The $a_1<a_2<a_3<...$'s are so chosen that the intervals will have no intersection.

Now, each $f_n$ integrates to 1 over $R$. However, the limit, which is the zero function, has an integral of zero.