Every positive r.v. X is the non-decreasing limit of a sequence
of simple r.v.’s. That is, if $X \geq 0$, then there exist simple r.v.’s
Xn, defined on the same probability triple, such that:
$\mathbf{x_n(\omega) \uparrow X(\omega) \quad \forall \quad \omega \in \Omega}$
The above proposition can be proved by constructing intervals
$\mathbf{A_jn=\{ \frac{j}{2^n} < X \leq \frac{j+1}{2^n} \} \quad for \quad j=0,1,2,...}$
Could anybody explain the steps? I cannot grasp the concept.
2026-04-06 02:41:53.1775443313
how to prove that every positive r.v. X is the non-decreasing limit of a sequence of simple r.v.’s
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Here is one standard construction:
Set $x_n(\omega) := 2^{-n} \lfloor (X(\omega)\wedge n) 2^n \rfloor$.
For the intuition: Imagine the area under the graph of a function $X: \mathbb{R} \to [0,\infty)$. At step $n$, you cut off everything above the level $n$, and then try to fill the space below the curve with vertical bars that have heights equal to multiples of $2^{-n}$. For example, in the following image, you see a curve that represents a measurable function $X$. In the first step, you are trying to fill everything below $1$ by chunks of height $1/2$ (big green chunks). In the second step, you are filling everything under $2$ by chunks with height $1/4$ (finer yellowish chunks). Then you fill everything up to the level $3$ with chunks of height $1/8$ (finer dark-red-brown part). In the forth step, you fill the area under $X \wedge 4$ with fine dark-red pieces. You proceed with bright red pieces to fill the tiny remaining gaps etc. The graphs that delimit the colored areas that you filled in $n$-th step are the $x_n$.