Is the condition "if $\phi\left(X_{n}\right)$ is integrable" redundant in this proposition?

Question

I'm reading a proposition about Martingale in my lecture note:

From the definition of martingale, I get $X_n$ is integrable. From this thread, I get the composition of a convex function with an integrable function is integrable. As such, $\phi(X_n)$ is automatically integrable.

Is the condition if $\phi\left(X_{n}\right)$ is integrable redundant?

Thank you so much ;)

Answer 1

If $X$ is integrable it does not follow that $X^{2}$ is. And $x \to x^{2}$ is convex. In the link you provided they are talking about Riemann integrability and boundedness of the function plays an important role. But an intergable random variable need not be bounded.

Answer 2

Of course not. Take for example $(\Omega ,\mathcal F,\mathbb P)=((0,1),\mathcal B,m)$ where $m$ is the Lebesgue measure and $\mathcal B$ being the $\sigma -$algebra of Borel set of $(0,1)$. Let $$X_n(\omega )=\omega,\quad \omega \in \Omega ,$$ for all $n$ and define $\varphi :(0,\infty )\to \mathbb R$ by $$\varphi (x)=\frac{1}{x}.$$ Then $\varphi $ is convex, but $\varphi (X_n)$ is never integrable.