Given a random variable $X$, if we take a measurable and bounded function $f(X)$ then can we say that $f$ is Lebesgue integrable wrt a probability measure on $\mathbb R$? In Real Analyses book by Rudin(1976) (pg315) the above is stated with the additional condition that the measure is finite. But a probability measure is always finite right? so we don't need to require that.
Now, the doubt that I have is why for some rv the $f(X)$ is not Lebesgue integrable (like the Cauchy case)? What I think is because usually $f$ is not bounded and non-integrability has nothing to do with the measurability or the fact that the probability measure is infinite.
So, my final question is weather is safe to say that if $f(X)$ is bounded and measurable then the Lebesgue integral $\int_{\mathbb R}f(X) \, dP$ is always defined? is that right?
Any references or hints are highly appreciated. Thanks
If $f$ is bounded then $f(X)$ will be Lebesgue-integrable even if $X$ is not. For example, if $X$ has a Cauchy distribution, then $\cos X$ has an expected value.
The reason is that bounded measurable functions on spaces of finite measure are integrable. The random variable $X$ is a measurable function on a probability space, whose total measure is $1$. The composition of two measurable functions is measurable, and if $f$ is bounded then so is $f(X)$. The random variable $f(X)$ is also a measurable function on a probability space with total measure $1$.