Can you bound this sequence of random variables? 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!


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 $Y \equiv F(X_{1})$.


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...
 A: 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 $Y\in L^1([0,1],\mathcal{B}, \mathbb{P})$.
