----- > [!proposition] Proposition. ([[approximation by simple functions]]) > > Suppose $(X, \Sigma)$ is a [[measure|measure space]] and $f:X \to [-\infty, \infty]$ a [[measurable function]]. Then there exists a [[sequence]] $f_{1},f_{2},\dots$ of [[simple function|simple]] [[measurable function|measurable functions]] $X \to \mathbb{R}$ with monotonically increasing pointwise magnitude [[pointwise convergence|converging pointwise]] to $f$, and this [[sequence|convergence]] is in fact [[uniform convergence|uniform]] should $f$ be [[bounded function|bounded]]. In other words, there exists a [[sequence]] $f_{1}f_{2},\dots$ of functions such that: > - Each $f_{k}$ is a [[simple function|simple]] [[measurable function]]. > - $|f_{k}(x)| \leq |f_{k+1}(x)|\leq |f(x)|$ for all $k \in \mathbb{N}$ and all $x \in X$. > - $\lim_{k \to \infty}f_{k}(x)=f(x)$ for every $x \in X$; > - $f_{1},f_{2},\dots$ [[uniform convergence|converges uniformly]] on $X$ to $f$ if $f$ is bounded. > > > ![[axler_simple_approx_custom_f_latex.gif]] > > [!note] > It is clear from the proof below that each nonzero $f_{k}(x)$ can be assumed to have the same sign as $f(x)$. E.g. if $f$ is nonnegative then it can be approximated by nonnegative simple functions. ^note > [!proof]- Proof. ([[approximation by simple functions]]) > > The animation above displays the gist of the proof idea: approximate $f$ by simple functions by quantizing to finer and finer intervals, clipped at each step so as to maintain simplicity. > > Specifically, for each index $k$ we subdivide each interval $[n,n+1)$, $n \in \mathbb{N}$, of $\mathbb{R}$ into $2^{k}$ equally sized half-open subintervals. Then we 'magnet down' $f_{k}(x)$, clipped at $k$, if $f(x) \geq 0$ and 'magnet up' $f_{k}(x)$, clipped at $-k$, if $f(x) <0$, that is, $f_{k}(x)=\begin{cases} > \frac{m}{2^{k}} & 0 \leq f(x)<k \text{ and } m \in \mathbb{Z} \text{ is such that } f(x) \in \left[ \frac{m}{2^{k}}, \frac{m+1}{2^{k}} \right) \\ > \frac{m+1}{2^{k}} & \text{if }-k < f(x) <0 \text{ and } m \in \mathbb{Z} \text{ is such that } f(x) \in \left[ \frac{m}{2^{k}}, \frac{m+1}{2^{k}} \right) \\ > k & \text{if }f(x) \geq k \\ > -k & \text{if }f(x) \leq -k. > \end{cases}$ > Each $f ^{-1}\left( [\frac{m}{2^{k}}, \frac{m+1}{2^{k}} \right))\in \Sigma$ because $f$ is a [[measurable function]]. Thus each $f_{k}$ is also a [[measurable function]]. Note that the claimed monotone magnitude property holds by construction. The definition of $f_{k}$ implies $|f_{k}(x)-f(x)| \leq \frac{1}{2^{k}}$ > for all $x \in X$ such that $f(x) \in [-k,k]$. [[Pointwise convergence]] follows, with [[uniform convergence]] if $f$ is [[bounded function|bounded]] (so that there exists $K$ such that $f(x) \in [-K,K]$ for all $x$). ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```