Bounded Measurable Function Uniform Limit of Simple Functions

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Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

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


Then there exists a sequence $\left({f_n}\right)_{n \in \N} \in \mathcal E \left({\Sigma}\right)$ of simple functions, such that:

$\forall \epsilon > 0: \exists n \in \N: \forall x \in X: \left\vert{f \left({x}\right) - f_n \left({x}\right)}\right\vert < \epsilon$

That is, such that $f = \displaystyle \lim_{n \to \infty} f_n$ uniformly.


The sequence $\left({f_n}\right)_{n \in \N}$ may furthermore be taken to satisfy:

$\forall n \in \N: \left\vert{f_n}\right\vert \le \left\vert{f}\right\vert$

where $\left\vert{f}\right\vert$ denotes the absolute value of $f$.


Proof

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

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

$A^n_k := \begin{cases} \left\{{ k 2^{-n} \le f < \left({k + 1}\right) 2^{-n} }\right\} & : k \ne n 2^n \\ \left\{{f \ge n}\right\} & : k = n 2^n \end{cases}$

where e.g. $\left\{{f \ge n}\right\}$ is short for $\left\{{x \in X: f \left({x}\right) \ge n}\right\}$.

It is immediate that the $A^n_k$ are pairwise disjoint, and that:

$\displaystyle \bigcup_{k \mathop = 0}^{n 2^n} A^n_k = X$

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

$f_n \left({x}\right) := \displaystyle \sum_{k \mathop = 0}^{n 2^n} k 2^{-n} \chi_{A^n_k} \left({x}\right)$

where $\chi_{A^n_k}$ is the characteristic function of $A^n_k$.


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

$x \in A^n_k$

so that:

\(\displaystyle \left\vert{f \left({x}\right) - f_n \left({x}\right)}\right\vert\) \(=\) \(\displaystyle \left\vert{f \left({x}\right) - k 2^{-n} }\right\vert\)
\(\displaystyle \) \(<\) \(\displaystyle 2^{-n}\)

since $x \in A^n_k$ if and only if $k 2^{-n} \le f \left({x}\right) < \left({k + 1}\right) 2^{-n}$.

In particular, since $f_n \left({x}\right) \le n$ 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:

$A^n_{n 2^n} = \left\{{f \ge n}\right\}$
$A^n_k = \left\{{f \ge k 2^{-n}}\right\} \cap \left\{{f < \left({k + 1}\right) 2^{-n}}\right\}$

are all $\Sigma$-measurable sets.

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


It remains to show that $\displaystyle \lim_{n \to \infty} f_n = f$ uniformly.

Let $\epsilon > 0$ be arbitrary.

Let $n \in \N$ be a bound for $f$ satisfying $2^{-n} < \epsilon$.

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

$\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert < 2^{-n}$


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 uniformly to $f^+$ and $f^-$, respectively.

That is, given $\epsilon > 0$, there exist $n, n'$ such that:

$\forall m \ge n: \forall x \in X: \left\vert{f^+ \left({x}\right) - f^+_m \left({x}\right)}\right\vert < \epsilon$
$\forall m \ge n': \forall x \in X: \left\vert{f^- \left({x}\right) - f^-_m \left({x}\right)}\right\vert < \epsilon$

By the Triangle Inequality, for all $m \in \N$:

$\left\vert{\left({f^+ \left({x}\right) - f^- \left({x}\right)}\right) - \left({f^+_m \left({x}\right) - f^-_m \left({x}\right)}\right)}\right\vert \le \left\vert{f^+ \left({x}\right) - f^+_m \left({x}\right)}\right\vert + \left\vert{f^- \left({x}\right) - f^-_m \left({x}\right)}\right\vert$

Hence, given $\epsilon > 0$, if $n, n'$ are sufficient for $\epsilon / 2$ for $f^+_m$ and $f^-_m$ respectively, it follows that for $m \ge \max \left\{{n, n'}\right\}$:

$\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert = \left\vert{\left({f^+ \left({x}\right) - f^- \left({x}\right)}\right) - \left({f^+_m \left({x}\right) - f^-_m \left({x}\right)}\right)}\right\vert < \epsilon$

where $f_m \left({x}\right) = f^+_m \left({x}\right) - f^-_m \left({x}\right)$.


Thus $f_m$ converges to $f$ uniformly.


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

$\left\vert{f^+_n \left({x}\right) - f^-_n \left({x}\right)}\right\vert = f^+_n \left({x}\right) + f^-_n \left({x}\right) \le f^+ \left({x}\right) + f^- \left({x}\right) = \left\vert{f \left({x}\right)}\right\vert$

where the last equality follows from Sum of Positive and Negative Parts.


Hence the result.

$\blacksquare$


Also see


Sources