Measurable Function is Pointwise Limit of Simple Functions

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Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.


Then there exists a sequence $\sequence {f_n}_{n \mathop \in \N} \in \map \EE \Sigma$ of simple functions, such that:

$\forall x \in X: \map f x = \ds \lim_{n \mathop \to \infty} \map {f_n} x$

That is, such that $f = \ds \lim_{n \mathop \to \infty} f_n$ pointwise.


If $f \ge 0$, the sequence $\sequence {f_n}_{n \mathop \in \N}$ may furthermore be taken to be increasing.


Proof

First, let us prove the theorem when $f$ is a positive $\Sigma$-measurable function.

Now for any $n \in \N$, define for $0 \le k \le n 2^n$:

${A_k}^n := \begin{cases}
\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
\set {f \ge n} & : k = n 2^n

\end{cases}$

where for example $\set {f \ge n}$ is short for $\set {x \in X: \map f x \ge n}$.

It is immediate that the ${A_k}^n$ are pairwise disjoint, and that:

$\ds \bigcup_{k \mathop = 0}^{n 2^n} {A_k}^n = X$

Subsequently, define $f_n: X \to \overline \R$ by:

$\map {f_n} x := \ds \sum_{k \mathop = 0}^{n 2^n} k 2^{-n} \map {\chi_{ {A_k}^n} } x$

where $\chi_{ {A_k}^n}$ is the characteristic function of ${A_k}^n$.


Now if $\map f x < n$, then we have for some $k < n 2^{-n}$:

$x \in {A_k}^n$

so that:

\(\ds \size {\map f x - \map {f_n} x}\) \(=\) \(\ds \size {\map f x - k 2^{-n} }\)
\(\ds \) \(<\) \(\ds 2^{-n}\)

since $x \in {A_k}^n$ if and only if $k 2^{-n} \le \map f x < \paren {k + 1} 2^{-n}$.

In particular, since $\map {f_n} x \le n$ for all $x \in X$, we conclude that pointwise, $f_n \le f$, for all $n \in \N$.


By Characterization of Measurable Functions and Sigma-Algebra Closed under Intersection, it follows that:

${A_{n 2^n} }^n = \set {f \ge n}$
${A_k}^n = \set {f \ge k 2^{-n} } \cap \set {f < \paren {k + 1} 2^{-n} }$

are all $\Sigma$-measurable sets.

Hence, by definition, all $f_n$ are $\Sigma$-simple functions.


It remains to show that $\ds \lim_{n \mathop \to \infty} f_n = f$ pointwise.

Let $x \in X$ be arbitrary.

If $\map f x = +\infty$, then for all $n \in \N$, $x \in {A_{n 2^n} }^n$, so that:

$\map {f_n} x = n$

Now clearly, $\ds \lim_{n \mathop \to \infty} n = +\infty$, showing convergence for these $x$.

If $\map f x < +\infty$, then for some $n \in \N$, $\map f x < n$.

By the reasoning above, we then have for all $m \ge n$:

$\size {\map f x - \map {f_m} x} < 2^{-m}$

which by Sequence of Powers of Number less than One implies that:

$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$

Thus $\ds \lim_{n \mathop \to \infty} f_n = f$ pointwise.


This establishes the result for positive measurable $f$.

For arbitrary $f$, by Difference of Positive and Negative Parts, we have:

$f = f^+ - f^-$

where $f^+$ and $f^-$ are the positive and negative parts of $f$.

By Function Measurable iff Positive and Negative Parts Measurable, $f^+$ and $f^-$ are positive measurable functions.

Thus we find sequences ${f_n}^+$ and ${f_n}^-$ converging pointwise to $f^+$ and $f^-$, respectively.

The Sum Rule for Real Sequences implies that for all $x \in X$:

$\ds \lim_{n \mathop \to \infty} \map { {f_n}^+} x - \map { {f_n}^-} x = \map {f^+} x - \map {f^-} x = \map f x$

Furthermore, we have for all $n \in \N$ and $x \in X$:

\(\ds \size {\map { {f_n}^+} x - \map { {f_n}^-} x}\) \(=\) \(\ds \map { {f_n}^+} x + \map { {f_n}^-} x \le \map {f^+} x + \map {f^-} x\)
\(\ds \) \(=\) \(\ds \size {\map f x}\) Sum of Positive and Negative Parts


This theorem requires a proof.
In particular: Prove that $\sequence {f_n}_{n \mathop \in \N}$ is increasing
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Hence the result.

$\blacksquare$


Also see


Sources

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