The problem is:
prove that space $L^q(\Omega,\mathcal{F},P)$ is contained in $L^r(\Omega,\mathcal{F},P)$ for any $1 \le r\le q$. where $P$ is probability measure
To prove this is sufficient to prove that $\|Y\|_r\le \|Y\|_q$
Then proof uses a truncated sequence $X_n = (\min(|Y|,n))^r$ to approximate $|Y|^r$,then using Jensen inequality on the truncated sequence. that is:
$(\mathbb{E}[X_n])^{\frac{q}{r}} \le \mathbb{E}[|Y|^q] $.
Then get's the result by monotonic convergence theorem
my question is why we use truncated sequence intead of using $|Y|^r$ replace of $X_n$ above?Is it because of Jensen inequality needs function integrable?
Is there some example show that random variable or convex function in Jensen inequality is not $L^1$ then it may not hold?
Best Answer
Well, assume your space is $([0,1],\mathcal{B},\lambda)$ with the latter denoting the Borel-algebra and the Lesbegue measure, and let $Y(x)=\frac{1}{x}$ and consider the convex function $f(x)=\frac{1}{x}$. Then, $f(Y(x))=x,$ which is clearly in $L^1$. However, $Y$ isn't.
Conversely, taking $f$ to be the same $Y(x)=x$ shows that the other way around might fail as well.
Thus, it is an important assumption that both $Y$ and $f(Y)$ are $L^1$-variables, since otherwise, either of the sides of the inequality might not even be defined.