Let $X_n$ be nonegative random variables in $(\Omega, F, \Bbb{P})$ and $F_n$ a sequence of sub-$\sigma$-algebras of $F$. Let me define $A_n:=\{\Bbb{E}(X_n|F_n)>\epsilon\}$ and denote $B_n:=A_n^c$. I want to check that $\Bbb{E}(X_n 1_{B_n})<\epsilon$.
I somehow get some problems in proving this rigorously. Intuitively it is clear to me. First I thought that I can work with $$\Bbb{E}(X_n 1_{B_n})=\int_{\Omega} X_n(\omega) 1_{B_n}(\omega)~d\Bbb{P}=\int_{B_n}X_n(\omega) ~d\Bbb{P}$$ but this do not helps me further. Then I thought that $$\Bbb{E}(X_n 1_{B_n})=\Bbb{E}\left(\Bbb{E}(X_n1_{B_n}|F_n)\right)\leq \Bbb{E}(\Bbb{E}(X_n|F_n))\stackrel{\text{in}~B_n}{\leq}\Bbb{E}(\epsilon)=\epsilon$$
But I also think that this is wrong.
It would be helpful if you could explain me how to do this but also maybe show where I did my mistake since then I can learn from this.
thanks a lot.
The only problematic step is $\mathbb E\left[\mathbb E\left[X_n\mid\mathcal F_n\right]\right]\leqslant\varepsilon$, since you lost the information of $B_n$. Instead, you can keep the indicator of $B_n$ and notice that this random variable is, by definition of the conditional expectation, $\mathcal F_n$-measurable: $$ \mathbb E\left[X_n\mathbf{1}_{B_n}\right]=\mathbb E\left[\mathbb E\left[X_n\mathbf{1}_{B_n}\mid\mathcal F_n\right]\right]=\mathbb E\left[\mathbf{1}_{B_n}\mathbb E\left[X_n\mid\mathcal F_n\right]\right] $$ and then use that $\mathbf{1}_{B_n}\mathbb E\left[X_n\mid\mathcal F_n\right]\leqslant\varepsilon$.
It seems that the best we can get is $ \mathbb E\left[X_n\mathbf{1}_{B_n}\right]\leqslant \varepsilon$ and that the strict inequality may not hold.