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This is part of a problem of showing that there exist a sequence of simple functions, $s_n$ that converges to a measurable function $g$.

For a fixed integer $n$, define $$ s_n(x) = \begin{cases} \frac{k}{2^n}, & \mbox{if } \frac{k}{2^n}\leq g(x) \lt \frac{k+1}{2^n}~,k=0,1,2,\ldots n2^n-1 \\ n, & \mbox{if } g(x)\geq n \end{cases} $$

I want to show that $s_1\leq s_2\leq s_3\leq \ldots $


On the interval $g(x) \geq n$, $s_{n+1} \geq n$. So on this interval, $s_{n+1} \geq s_n$.

If $g(x) < n$ , then $\frac{k}{2^n}\leq g(x) \lt \frac{k+1}{2^n},k=0,1,2,\ldots n2^n-1 $. I need some help with this part.


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1 Answer 1

up vote 3 down vote accepted

Suppose $g(x)<n$. Then $s_n(x)={k\over 2^n} $ for some $0\le k\le n2^n-1$ and $$\tag{1}{k\over 2^n}\le g(x)<{k+1 \over 2^n}.$$ Now, for the next "level of sets" for $s_{n+1}$, we have $$\textstyle\bigl[\,{k\over 2^n},{k+1 \over 2^n} \,\bigr) =\bigl[\,{2k\over 2^{n+1}},{2k+1 \over 2^{n+1}}\,\bigr ) \cup\bigl[\,{2k+1\over 2^{n+1}},{2k+2 \over 2^{n+1}} \bigr).$$

We know from $(1)$ that $g(x)$ is in either the set $\bigl[\,{2k\over 2^{n+1}},{2k+1 \over 2^{n+1}}\,\bigr )$ or in the set
$\bigl[\,{2k+1\over 2^{n+1}},{2k+2 \over 2^{n+1}}\,\bigr )$; so $s_{n+1}(x)$ either takes the value ${2k\over 2^{n+1}}={k\over 2^n}$ or the value ${2k+1\over 2^{n+1}}>{k\over 2^n}$.

Thus $s_{n+1}(x)\ge s_n(x)$ whenever $g(x)<n$.

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So the key step is recognizing that $$\textstyle\bigl[\,{k\over 2^n},{k+1 \over 2^n} \,\bigr) =\bigl[\,{2k\over 2^{n+1}},{2k+1 \over 2^{n+1}}\,\bigr ) \cup\bigl[\,{2k+1\over 2^{n+1}},{2k+2 \over 2^{n+1}} \bigr).$$? –  Kuku Feb 27 '12 at 18:54
@kuku Yes, exactly. The author uses the so-called dyadic intervals. The "next stage" of intervals is obtained by splitting each interval in the previous stage in half (plus he adds more intervals "at the top"). –  David Mitra Feb 27 '12 at 19:00

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