Take measure space as $\left(X,\mathcal B (X),\mu \right)$ where $\mathcal B (X)$ is Borel $\sigma$-algebra of $X$ and $\mu$ is Lebesgue measure.
Show that $\int fd\mu \neq \lim_{n\to \infty} \int f_n d\mu$ where $f_n= \frac1n$ and $f=0$
I have shown that $f_n$ is decreasing sequence and converges to $f$ uniformly and $\int fd\mu =0 $ (a.e.) but I cannot identify what is $\int f_n d\mu$
Our definition for positive functions is $\int fd\mu = \sup \{\varphi : 0 \leq \varphi \leq f \ \ \text{and} \ \ \varphi \ \ \text{is simple} \}$
I appreciate any help.
Thanks in advance
Assume that the space has infinite measure.
If $\forall n \in \Bbb{N},$you take $\phi_n(x)=\frac{1}{n}1_X$ then $0 \leq \phi_n \leq f$ and
$\int_Xf_n \geq \frac{1}{n}\mu(X)$ by definition.
So the limit is not true.
In a space of finite measure,it is true .