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simple function逼近measurable function

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hbghlyj Posted at 2023-9-24 12:11:45 |Read mode
Last edited by hbghlyj at 2023-11-8 14:47:00Theorem 4.1 Suppose $f$ is a non-negative measurable function on $\mathbb{R}^d$. Then there exists an increasing sequence of non-negative simple functions $\left\{\varphi_k\right\}_{k=1}^{\infty}$ that converges pointwise to $f$, namely,
$$
\varphi_k(x) \leq \varphi_{k+1}(x) \quad \text { and } \quad \lim _{k \rightarrow \infty} \varphi_k(x)=f(x), \text { for all } x .
$$

Proof. We begin first with a truncation. For $N \geq 1$, let $Q_N$ denote the cube centered at the origin and of side length $N$. Then we define
$$
F_N(x)=\left\{\begin{array}{cl}
f(x) & \text { if } x \in Q_N \text { and } f(x) \leq N \\
N & \text { if } x \in Q_N \text { and } f(x)>N \\
0 & \text { otherwise. }
\end{array}\right.
$$

Then, $F_N(x) \rightarrow f(x)$ as $N$ tends to infinity for all $x$. Now, we partition the range of $F_N$, namely $[0, N]$, as follows. For fixed $N, M \geq 1$, we define
$$
E_{\ell, M}=\left\{x \in Q_N: \frac{\ell}{M}<F_N(x) \leq \frac{\ell+1}{M}\right\}, \quad \text { for } 0 \leq \ell<N M
$$

Then we may form
$$
F_{N, M}(x)=\sum_{\ell} \frac{\ell}{M} \chi_{E_{\ell, M}}(x)
$$

Each $F_{N, M}$ is a simple function that satisfies $0 \leq F_N(x)-F_{N, M}(x) \leq$ $1 / M$ for all $x$. If we now choose $N=M=2^k$ with $k \geq 1$ integral, and let $\varphi_k=F_{2^k, 2^k}$, then we see that $0 \leq F_M(x)-\varphi_k(x) \leq 1 / 2^k$ for all $x$, $\left\{\varphi_k\right\}$ is increasing, and this sequence satisfies all the desired properties.

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Czhang271828 Posted at 2023-9-24 12:38:28

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