Can you bound this sequence of random variables?

Question

Let $(\Omega,\mathcal{F}, \mathbb{P})$ be a probability space and $\{X_{n}\}$ a decreasing family of non-negative random variables, such that $\{X_{n}\}\downarrow X$ almost surely. Moreover, let $F:[0,\infty)\rightarrow [0,\infty)$ be continuous with $F(X_{n})\in L^1(\Omega,\mathcal{F}, \mathbb{P})$ for all $n$.

Is the sequence $\{F(X_{n})\}$ bounded by some $Y\in L^1(\Omega,\mathcal{F}, \mathbb{P})$?

Intuitively, continuity of $F$ on $[0,\infty)$ should imply boundedness (see update below), but I am missing a good starting point.

Any comments, ideas and suggestions are highly appreciated. Thank you!

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EDIT

Please note that for any non-decreasing (measurable, but not necessarily continuous) $F$, the question can be answered affirmatively. In this case:

$$X_{n} \;\leq\; X_{1} \quad\Rightarrow\quad F(X_{n}) \;\leq\; F(X_{1})$$

and, because $F(X_{1})\in L^1(\Omega,\mathcal{F}, \mathbb{P})$ by assumption, we can simply set $Y \equiv F(X_{1})$.

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UPDATE

I should probably state my intuition more clearly. The "natural bound" for the random variables $F(X_{n})$ is

$$Y(\omega) = \max\;\{\, F(x) \;:\; x\in[\, 0,\, X_{1}(\omega) \,] \,\} \;< \infty$$

which is actually measurable (by the measurable selection theorem) and therefore a valid random variable. I don't see any reason why $Y$ should not be integrable. But then again, I also can't show that it is...

Answer 1

Is the sequence $\left\{F(X_{n})\right\}$ bounded by some $Y\in L_{1}(\Omega,\mathcal{F},\mathbb{P})$?

The answer is no, that is, in general there is no random variable $ Y\in L_{1}(\Omega,\mathcal{F},\mathbb{P})$ such that it is a bound of the sequence $\left\{F(X_{n})\right\}$.

The reason is that as $\{X_{n}\}\downarrow X$ almost surely then $\{F(X_{n})\}\downarrow F(X)$ almost surely, if there were such a function $ Y\in L_{1}(\Omega,\mathcal{F},\mathbb{P})$ then by The dominated convergence theorem we would have $\{F(X_{n})\}\downarrow F(X)$ in $L_{1}$, but, in general, we know that $\mathbb{P}$ almost surely convergence no implies $L_{1}$ convergence.

For the answer to your question to be true you should add conditions to $ X_ {n}$, $ X $ and $ f $.

Example: We consider the probability space $([0,1],\mathcal{B},\mathbb{P})$ where $\mathcal{B}$ is borel $\sigma$-algebra in $[0,1]$ and $\mathbb{P}$ is the probability measure given by $$ \mathbb{P}(A):=\left\{\begin{array}{ll} 1 &\mbox{ if }A=[0,1] \\ 0 & \mbox{ if }A\neq[0,1] \end{array}\right. $$ We consider the random variables $X_{n}:[0,1]\rightarrow \mathbb{R}$ defined by $$X_{n}(\omega):=\left\{\begin{array}{ll} n &\mbox{ if }\omega=\frac{1}{n} \\ 0 & \mbox{ if }\omega\neq\frac{1}{n} \end{array}\right. . $$ Note that $X_{n}\rightarrow 0$ almost surely. Now, consider $F=\mathbf{id}$ (identity), this is clearly continuous, futhermore, $F(X_{n})=X_{n}$, therefore $$\int_{[0,1]} F(X_{n})d\mathbb{P}=\int_{[0,1]} X_{n}d\mathbb{P}=n\mathbb{P}\left(X_{n}=n\right)=n\mathbb{P}\left(\left\{\frac{1}{n}\right\}\right)=0<\infty.$$ Then $F(X_{n})\in L^1([0,1],\mathcal{B}, \mathbb{P}) $. But clearly the sequence $\left\{F(X_{n})\right\}$ is not bounded by some $Y\in L^1([0,1],\mathcal{B}, \mathbb{P})$.