Disclaimer: This might be very trivial, but I haven't learned any of this formally. (I didn't know what a Lebesgue integral was before looking it up today.) It is possible I am not understanding the definitions correctly, which is part of the reason I'm asking.
Context: I have a finite sample space $\Omega$ and a measure $\pi : \Omega \to \mathbb{R}$. (My event set, per the formal definition of a measure space, is trivial in the sense that it is just the power set of $\Omega$. Thus, I'm just keeping it implicit.)
I have a bounded function $f : \Omega \to \mathbb{R}$, and I want to check whether it is in $L^2(\Omega, \pi)$. I have theorized that since $f$ is bounded, it immediately follows that $f \in L^p(\Omega, \pi)$ for all $p \ge 1$. (So in particular, $f \in L^2(\Omega, \pi)$.)
My thought process: I am working in the context of a finite probability space, but since I think my hypothesis holds even if this isn't true, let's work in a more general setting. Let $(X, \mathcal{F}, \mu)$ denote a $\sigma$-finite measure space and let $1 \le p < \infty$. According to my digging, $f \in L^p(X, \mu)$ if
$$
\int |f|^p d \mu < \infty.
$$
Per the definition of Lebesgue integration, this is equivalent to checking whether
$$
\int_{0}^\infty f^*(t) dt < \infty,
$$
where $f^*(t) = \mu(\{ x | x \in X, |f(x)|^p > t\})$. (We have converted the Lebesgue integral to an ordinary improper Riemann integral.)
Since $f$ is bounded, it follows that $|f|^p$ is bounded. This means there is some $C \ge 0$ such that $f^*(t) = 0$ for all $t \ge C$. So
$$
\int_{0}^\infty f^*(t) dt = \int_{0}^C f^*(t) dt.
$$
We know that $f^*(t)$ is bounded since the measure space is $\sigma$-finite. (I guess if it wasn't $\sigma$-finite than my hypothesis wouldn't hold, although I don't know why you would care about such a scenario.) It follows that
$$
\int_{0}^C f^*(t) dt
$$
is finite, and thus $f \in L^p(X, \mu)$.
(I saw that formally $L^p$ spaces are equivalence classes of functions, but I'm ignoring this formalism since it doesn't seem to affect my argument.)
Results from searching: This says that any bounded function on $[0, 1]$ is automatically $L^p$ for every value of $p$. But I don't see why we need to restrict ourselves to $[0, 1]$.
Thanks for any help!
Edit: I realize now that when I wrote this question I had the definition of "$\sigma$-finite measure" confused with "finite measure." (See the accepted answer.) I also realized that you have to be more careful when taking the Lebesgue integral of signed functions, so let's retroactively say that $f$ is nonnegative. (My bounds on the improper Riemann integral incorrectly went from $-\infty$ to $\infty$ before; now they correctly go from $0$ to $\infty$.)
Best Answer
This is true if and only if ($f$ is measurable and) the total measure is finite (this is where the restriction to $[0, 1]$ is from; it's false for $\mathbb{R}$ with the Lebesgue measure, for example). Otherwise not even the constant function $1$ is in $L^p$ for $p \in [1, \infty)$. Your argument implicitly assumes this when you say that $f^{\ast}$ is bounded (which implies finite) even though you only say you assume the measure is $\sigma$-finite.
If the total measure $\mu(X)$ is finite and $|f|$ is bounded by a constant $C$ then
$$\int_X |f|^p \, d \mu \le C^p \mu(X).$$
If the sample space is finite then $f$ is automatically measurable and the total measure is automatically finite so there's no issue there; in this very special case every function is (measurable and) bounded and so every function is in $L^p$ for every $p$.