$L^1$ random variable property

Question

If $X\in L^1$ is a random variable, then $$\forall\epsilon>0:\exists\delta>0:\forall \operatorname{event}A:\left(P(A)\leq\delta\Rightarrow\mathbb{E}(|X| 1_A)\leq\epsilon\right)\,.$$

Could anybody give me a hint as to how to (dis)prove this?

Answer 1

By definition of the Lebesgue integral, for any positive $\varepsilon$, there exists $Y$ such that

1. $0\leqslant Y\leqslant \left|X\right|$,
2. there exist an integer $N$, constants $a_1,\dots,a_N$ and measurable sets $B_1,\dots,B_N$ such that $Y=\sum_{i=1}^Na_i\mathbf 1_{B_i}$, where $\mathbf 1$ denotes the indicator function of a set,
3. $\mathbb E\left|X\right|-\mathbb E\left[Y\right]\lt \varepsilon/2$.

With this in mind, note that $$\tag{*} \mathbb E\left[|X|\mathbf 1_A\right]=\mathbb E\left[\left(|X|-Y\right)\mathbf 1_A\right]+\mathbb E\left[Y\mathbf 1_A\right]$$
and by 1., the first term of the right hand side does not exceed $\mathbb E\left[ |X|-Y \right]$ which can be in turn bounded by $\varepsilon/2$, by 3. The second term of the right hand side of (*) is smaller that $M\mathbb P\left(A\right)$, where $M=\max_{1\leqslant i\leqslant n}a_i$. In conclusion, we are sure that for any measurable set $A$, $$\mathbb E\left[|X|\mathbf 1_A\right]\leqslant \varepsilon/2+M\mathbb P\left(A\right).$$ Do you see which choice of $\delta$ will work?