Measurable Function Pointwise Limit of Simple Functions
Contents
Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.
Let $f: X \to \overline{\R}$ be a $\Sigma$-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:
- $\forall x \in X: f \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right)$
That is, such that $f = \displaystyle \lim_{n \to \infty} f_n$ pointwise.
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 $\displaystyle \lim_{n \to \infty} f_n = f$ pointwise.
Let $x \in X$ be arbitrary.
If $f \left({x}\right) = +\infty$, then for all $n \in \N$, $x \in A^n_{n 2^n}$, so that:
- $f_n \left({x}\right) = n$
Now clearly, $\displaystyle \lim_{n \to \infty} n = +\infty$, showing convergence for these $x$.
If $f \left({x}\right) < +\infty$, then for some $n \in \N$, $f \left({x}\right) < n$.
By the reasoning above, we then have for all $m \ge n$:
- $\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert < 2^{-m}$
which by Sequence of Powers of Number less than One implies that:
- $\displaystyle \lim_{n \to \infty} f_n \left({x}\right) = f \left({x}\right)$
Thus $\displaystyle \lim_{n \to \infty} f_n = f$ 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 $x \in X$:
- $\displaystyle \lim_{n \to \infty} f^+_n \left({x}\right) - f^-_n \left({x}\right) = f^+ \left({x}\right) - f^- \left({x}\right) = f \left({x}\right)$
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
- Bounded Measurable Function Uniform Limit of Simple Functions, a strengthening when $f$ is bounded
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.8$