The problem statement is
If $\mu(X) < +\infty$ and $f \in L^2(\mu)$ on $X$, prove that $f \in L(\mu)$ on $X$. If $\mu(X) = \infty$, this is false.
I came up with a solution, but it doesn't match what's in the solutions manual, so I want to see why it's wrong. I thought that maybe I could use a characteristic function and theorem 11.35 to show this. Here is my argument:
Take $K_X$ to be the characteristic function on $X$, so that $K_X(x) = 1$ if $x\in X$ and $0$ otherwise. It will be shown that $K_X \in L^2(\mu)$ on $X$. Note that $|1|^2 = 1$ and $|0|^2 = 0$, which implies that $|K_X|^2 = K_X$. But, then $$\int_X |K_X|^2 d\mu = \int_X K_X d\mu = \mu(X) < +\infty. $$ It follows from definition 11.34 that $K_X \in L^2(\mu)$ on $X$. By applying theorem 11.35, it follows that $K_X f \in L(\mu)$ on $X$. But, on the set $X$, we have $K_X f = f$. Therefore, $f\in L(\mu)$, and we have the claim.
I think this is okay, but I guess I'm not sure if it's valid to say that $K_Xf = f$ on the set $X$. Can anyone point me in the right direction?
edit: updating with definitions.
Definition 11.34: Let $X$ be a measurable space. We say that $f\in L^2(\mu)$ on $X$ if $f$ is measurable and if $$\int_X |f|^2 d\mu < +\infty$$.
Theorem 11.35: Suppose that $f \in L^2(\mu)$ and that $g \in L^2(\mu)$. Then $fg \in L(\mu)$ and $$ \int_X |fg| d\mu \leq ||f|| ||g||$$. (The schwarz inequality).
We say that $f\in L(\mu)$ if $\int_X f d\mu = \int_X f^+ d\mu - \int_X f^- d\mu$, and both terms on the right hand side are finite. Here $f^+ = \max\{f, 0\}$ and $f^- = -\min \{f, 0\}$
Hint: Write $$f(x) = 1_{|f(x)| \geq 1}(x) f(x) + 1_{|f(x)| \leq 1}(x) f(x).$$