# Does $f_n \to f$ converges in $L^p$ imply exchangeability of integral and limit on $f_n$?

If $f_n \to f$ converges in $L^p$, does $\int f d\mu = \lim_n \int f_n d \mu$?

When $p=1$, $$- \lim_n \int |f_n -f| d \mu \leq \lim_n \int (f_n -f) d \mu \leq \lim_n \int |f_n -f| d \mu$$ So $\int f d\mu = \lim_n \int f_n d \mu$.

But what if $p> 1$ and $p \in (0,1)$? Thanks!

-

## 1 Answer

Take $p = 2$, $f_n(x) = \dfrac{1}{nx}$ and $f(x) = 0$.

$$\|f_n - f\|_2 = \left\{\int_1^\infty \left|\dfrac{1}{nx}\right|^2\,dx\right\}^{1/2} = \dfrac{1}{n}$$

Therefore, $f_n \to f$ in $L^2([1, \infty])$. However:

$$\int_1^\infty f_n(x) \,dx = \infty \quad \forall n$$

You can construct a similar counterexample on $L^{1/2}([0, 1])$.

-