Definition: 令 $f_n, f$ 為 complex-valued measurable 函數，我們說 $f_n \to f$ almost uniformly 若下列條件成立：
對任意 $\varepsilon>0$，存在可測集 $E_\varepsilon$ 滿足 $\mu(E_\varepsilon)<\varepsilon$ 使得
\[
f_n \to f \text{ uniformly on $E_\varepsilon^c$}
\]換言之，$\sup_{x \in X\setminus E_\varepsilon} |f_n(x) - f(x)| < \varepsilon$。
Example:
考慮 測度空間 $([0,1],\mathcal{L},m)$，取 $f_n(x) := x^n$ ，則
1. $f_n \not \to f$ uniformly
2. $f_n \to f$ almost uniformly
Proof 1.:
首先注意到 $$
\lim_n f_n(x) := f(x) = \begin{cases}
0 & x \in [0,1) \\
1 & x = 1
\end{cases}
$$ 觀察 $$
\sup_{x \in [0,1]} |f_n(x) - f(x)| = 1 \not\to 0
$$故 $f_n \not\to f$ uniformly。$\square$
Proof 2: 令 $\varepsilon \in (0,1)$ 取 $E_\varepsilon :=(1-\varepsilon/2, 1]$ 則 $\mu(E_\varepsilon) = \varepsilon/2 < \varepsilon$ 且對任意 $x \in [0,1] \setminus E_\varepsilon$ 而言，
\begin{align*}
\sup_{x \in [0,1]\setminus E_\varepsilon} |f_n(x) - f(x)| &= \sup_{x \in [0, 1-\varepsilon/2]}|x^n-0|\\
&=(1-\varepsilon/2)^n \to 0 \;\; \text{ as $n \to \infty$}
\end{align*}
至此證明完畢。$\square$
對任意 $\varepsilon>0$，存在可測集 $E_\varepsilon$ 滿足 $\mu(E_\varepsilon)<\varepsilon$ 使得
\[
f_n \to f \text{ uniformly on $E_\varepsilon^c$}
\]換言之，$\sup_{x \in X\setminus E_\varepsilon} |f_n(x) - f(x)| < \varepsilon$。
Example:
考慮 測度空間 $([0,1],\mathcal{L},m)$，取 $f_n(x) := x^n$ ，則
1. $f_n \not \to f$ uniformly
2. $f_n \to f$ almost uniformly
Proof 1.:
首先注意到 $$
\lim_n f_n(x) := f(x) = \begin{cases}
0 & x \in [0,1) \\
1 & x = 1
\end{cases}
$$ 觀察 $$
\sup_{x \in [0,1]} |f_n(x) - f(x)| = 1 \not\to 0
$$故 $f_n \not\to f$ uniformly。$\square$
Proof 2: 令 $\varepsilon \in (0,1)$ 取 $E_\varepsilon :=(1-\varepsilon/2, 1]$ 則 $\mu(E_\varepsilon) = \varepsilon/2 < \varepsilon$ 且對任意 $x \in [0,1] \setminus E_\varepsilon$ 而言，
\begin{align*}
\sup_{x \in [0,1]\setminus E_\varepsilon} |f_n(x) - f(x)| &= \sup_{x \in [0, 1-\varepsilon/2]}|x^n-0|\\
&=(1-\varepsilon/2)^n \to 0 \;\; \text{ as $n \to \infty$}
\end{align*}
至此證明完畢。$\square$
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