📂Measure Theory

Existence of a Sequence of Simple Functions Converging to a Measurable Function

Theorem¹

Let $(X, \mathcal{E})$ be a measurable space.

1. If $f : X \to [0, \infty]$ is a measurable function, then there exists a sequence of simple functions $\left\{ \phi_{n} \right\}$ satisfying the following: $$ 0 \le \phi_{1} \le \phi_{2} \le \cdots \le f \quad \text{and} \quad \phi \to f $$ If $f$ is bounded, $$ \phi \rightrightarrows f $$

Here, $\phi \to f$ denotes pointwise convergence, and $\phi \rightrightarrows f$ denotes uniform convergence.

2. If $f : X \to \mathbb{C}$ is a measurable function, then there exists a sequence of simple functions $\left\{ \phi_{n} \right\}$ satisfying the following: $$ 0 \le \left| \phi_{1} \right| \le \left| \phi_{2} \right| \le \cdots \le \left| f \right| \quad \text{and} \quad \phi \to f $$ If $f$ is bounded, $$ \phi \rightrightarrows f $$

Proof

We prove for real functions only.

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For $n = 0, 1, 2, \dots$ and $0 \le k \le 2^{2n} -1$, let $E_{n}^{k}$ and $F_{n}$ be as follows.

$$ E_{n}^{k} = f^{-1}\left( (k2^{-n}, (k+1)2^{-n}] \right) \quad \text{ and } \quad F_{n} = f^{-1}\left( (2^{n}, \infty] \right) $$

And let $\phi_{n}$ be defined as below.

$$ \phi_{n} = \sum\limits_{k=0}^{2^{2n} -1}k2^{-n}\chi_{E_{n}^{k}} + 2^{n}\chi_{F_{n}} $$

$\chi$ is a characteristic function. It might be hard to understand with formulas alone, so look at the picture below.

The left picture shows some $f$ and $phi_{0}$, and the right picture shows $f$ and $\phi_{1}$. It will be easier to understand how $\phi_{n}$ is constructed if we consider it one by one from $n=0$.

Then, by definition, $\phi_{n} \le \phi_{n+1}$ holds. Also, when $f \le 2^{n}$, $f - \phi_{n} \le 2^{-n}$ holds (see the picture above).

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