Convergence in $p$th mean does not imply convergence in mean. My book proves the following:
Let $(\Omega,\mathcal{A},\mu)$ be a finite measure space. Then every sequence $(f_n)$ in $\mathcal{L}^p$ which converges in $p$th mean to an $f \in \mathcal{L}^p$ for some $p\geq 1$ also converges to $f$ in mean.
and then ask to provide an example where it fails when $\mu$ is not finite. It hints me to the following previous example:
$\Omega=\mathbb{N}$, $\mathcal{A}=\mathcal{P}(\mathbb{N})$, and $\mu$ defined by $\alpha_n:=\mu(\{n\})=\frac{1}{\sqrt{n}}$ for each $n\in\mathbb{N}$. Then if $f$ is the function on $\Omega$ defined by $f(n)=\alpha_n$ for each $n$ we see that $f\in\mathcal{L}^2$ but $f\notin\mathcal{L}^1$.
Am having trouble using this to construct the aforementioned example. Any help is greatly appreciated.
 A: We have $f \in \mathcal{L}^2(\Omega)$ because
$$
\int f^2 d\mu = \sum_{n = 1}^{\infty} f(n)^2 \mu(\{n\}) = \sum_{n = 1}^{\infty} \frac1n\cdot \frac1{\sqrt{n}} < \infty.
$$
We have $f\not\in \mathcal{L}^1(\Omega)$ because
$$
\int f d\mu = \sum_{n = 1}^{\infty} f(n) \mu(\{n\}) = \sum_{n = 1}^{\infty} \frac1{\sqrt{n}}\cdot \frac1{\sqrt{n}} = \sum_{n = 1}^{\infty} \frac1n = \infty.
$$
Now define for $m \in \mathbb N$ the function $g_m:\Omega \rightarrow \mathbb R$ by
$$
g_m(n) = f(n)\quad if \quad n < m,\qquad g_m(n) = 0\quad if \quad n \geq m.
$$
We find
$$
\int (f - g_m)^2 d\mu = \sum_{n = m}^{\infty} f(n)^2 \mu(\{n\}) = \sum_{n = m}^{\infty} \frac1n\cdot \frac1{\sqrt{n}} \rightarrow 0 \quad for \quad m \rightarrow \infty,
$$
and hence
$$
g_m \rightarrow f \quad in \quad \mathcal{L}^2(\Omega).
$$
We also find
$$
\int (f - g_m) d\mu = \sum_{n = m}^{\infty} f(n) \mu(\{n\}) = \sum_{n = m}^{\infty} \frac1n \not\rightarrow 0 \quad for \quad m \rightarrow \infty,
$$
and hence
$$
g_m \not\rightarrow f \quad in \quad \mathcal{L}^1(\Omega).
$$
In fact, it's not so difficult to check along the lines of the calculations above that the sequence $g_m$ is not Cauchy in $\mathcal{L}^1(\Omega)$, and hence does not converge at all in $\mathcal{L}^1(\Omega)$.
A: For $1<p<\infty,$ we can use Holder:
$$\int_X|f_n-f|\,d\mu = \int_X|f_n-f|\cdot 1\,d\mu$$ $$ \le (\int_X|f_n-f|^p\,d\mu)^{1/p}\cdot (\int_X 1^q\,d\mu)^{1/q} $$ $$= (\int_X|f_n-f|^p\,d\mu)^{1/p}\mu(X)^{1/q} \to 0.$$
An example where $\mu(X)=\infty$ and the result fails: On $[1,\infty)$ with Lebesgue measure, define
$$f_n(x) = \frac{1}{nx}.$$
Then for $1<p<\infty,$
$$\int_1^\infty |f_n(x)-0|^p\, dx = \frac{1}{n^p(p-1)}\to 0.$$
But if $p=1,$
$$\int_1^\infty |f_n(x)-0|\, dx =\infty$$
for all $n,$ so there is no hope.
Note that I've left out $p=\infty.$ You should try this case on your own at first.
