Let $Z \in L_1(\Omega, \Sigma, \mu)$.
Let $X=\max(\min(k, Z), -k)$ be absolutely bounded, so $|X| < k$.
Does this imply that $X \in L_2(\Omega, \Sigma, \mu)$?
What if $X$ is uniformly continuous?
I am trying to understand a proof and this is a property I do not understand - the author claims $X$ is in $L_2$ with the aforementioned properties, but I am not sure if this is the case.
I know of course that boundedness is not enough for a function to be in $L_2$ for arbitrary $\Omega$, but if it is already in $L_1$ some properties might hold that make it true. I was thinking that since it's already in $L_1$ it must be asymptotically approaching zero fast enough at the borders of $\Omega$ and if it has singularities at non-border points they will get "cut" by the bounds, so they should theoretically integrate to a finite value. I think it does make sense, but I do not know how to prove it.
If this is not true, what are the assumptions on $X$ or $\Omega$ that can make it true?
This follows from Holder's inequality:
$$ \| f \|_2^2 = \| f^2 \|_1 \leq \| f \|_1 \| f \|_\infty < \infty $$