Approximate $X$ by a sequence of simple random variables $X_n$ in $l_1$ norm such that $|X_n|<|X|$ Given a random variable $X$ with finite expectation, I know that $$X_n\to X, a.s.\text{and}  |X_n| \leq X\implies \mathbb{E}|X-X_n|\to 0 \text{ by DCT.}$$
I am wondering if it is possible to approximate $X$ (with finite expectation) by a sequence of simple random variables:
$$\forall \varepsilon, \exists \text{  a simple r.v.  } X_{{\varepsilon}} \text{ such that } |X_{{\varepsilon}}|\leq |X| \text{ and } \mathbb{E}|X-X_{{\varepsilon}}|< \varepsilon.$$
Any help will be greatly appreciated!
 A: Decompose $X$ as $X=X^+-X^-$, where $X^+=\max\{X,0\}$ and $X^-=\max\{-X,0\}$.
By definition of Lebesgue integral, there exist simple non-negative random variables $X_\varepsilon^+$ and $X_\varepsilon^-$ such that $X_\varepsilon^+\leqslant X^+$, $X_\varepsilon^-\leqslant X^-$ and $\max\{\mathbb E\left[X^+-X_\varepsilon^+\right],\mathbb E\left[X^--X_\varepsilon^-\right]\}\lt \varepsilon/2$.
Let $X_\varepsilon:=X_\varepsilon^+-X_\varepsilon^-$; then $X_\varepsilon$ is a simple random variable. Observe that
$$
X_\varepsilon\leqslant X_\varepsilon^+\leqslant X^+\leqslant X
$$
and
$$
-X_\varepsilon\leqslant X_\varepsilon^-\leqslant X^-\leqslant -X
$$
hence $\lvert X_\varepsilon\rvert\leqslant \lvert X\rvert$. Moreover,
$$
\mathbb E\lvert X-X_\varepsilon\rvert\leqslant \mathbb E\lvert X^+-X^+_\varepsilon\rvert+\mathbb E\lvert X^--X^-_\varepsilon\rvert\lt\varepsilon.
$$
A: At least for positive random variables it is very close to how the Lebesgue integral (expectation) is defined. Note that if $(\Omega,\mathcal{F},P)$ is the underlying probability space, then
$$\mathbb{E}X = \int_{\Omega} X(\omega) dP(\omega).$$
If $X\geq 0$ is measurable, then the integral of $X$ is defined as the supremum over the simple functions which approximate $X$ from below. An approximating sequence can be constructed as follows.
Let $n\in \mathbb{N}$ and consider a simple function:
$$\begin{cases} X_n(\omega) = n,\quad &{\rm if }\ X(\omega)\geq n,\\ X_n(\omega) = 2^{-n}(k-1),\quad &{\rm if} \ X(\omega)\in [2^{-n}(k-1),2^{-n}k),\ k=1,2,\ldots,n2^n. \end{cases}$$
We have $X_n\nearrow X$ almost everywhere and the convergence of integrals follows from the monotone convergence theorem.
Usually in the definition of the integral for a signed function $X$, you decompose into the positive and the negative part: $X = X_+ - X_-$. If at least one of the functions $X_+$, $X_-$ is integrable, then the integral of $X$ makes sense as the difference of the two integrals. Knowing that $|X|$ has finite expectation we actually know that both $X_+$ and $X_-$ have finite first moments. Thus we may slightly modify the above construction for the non-negative functions in order to obtain an approximating sequence $|X_n|\leq |X|$ for signed r.v. $X$.
