Lemma 5

Let $X$ be a locally compact space, $\mu$ a bounded positive measure on $X$, $F$ a Banach space, and $f$ a bounded function on $X$ with values in $F$. The following conditions are equivalent:

(i) The set of points of discontinuity of $f$ is $\mu$-negligible.

(ii) For every $\varepsilon > 0$, there exist elements $a_1, \ldots, a_n$ of $F$, functions $g_1, \ldots, g_n$ belonging to $\mathscr{K}(X)$, and a bounded continuous function $h \geq 0$ on $X$ such that $|f - g_1 a_1 - \cdots - g_n a_n| \leq h \leq 2 \sup |f|$ on $X$, and $\int h d\mu \leq \varepsilon$.

Denote by $N$ the set of points of discontinuity of $f$, and let $M = \sup |f|$.

(i) $\Rightarrow$ (ii). Suppose that condition (i) is satisfied. Let $\varepsilon > 0$. The function $f$ is $\mu$-integrable (No. 2, Cor. 4 of Prop. 5, and No. 6, Th. 5), therefore there exist $a_1, \ldots, a_n$ in $F$ and $g_1, \ldots, g_n$ in $\mathscr{K}(X)$ such that, on setting $k = |f - g_1 a_1 - \cdots - g_n a_n|$, we have $\int k d\mu \leq \varepsilon/2$ (§ 3, No. 5,

Prop. 10). Multiplying $g_1, \ldots, g_n$ by a suitable same element of $\mathscr{K}(X)$, we can suppose in addition that

on X, whence $k \leq 2M$. The set $N'$ of points of discontinuity of $k$ is contained in $N$, hence is negligible. For every $x \in X$, set $l(x) = \limsup_{y \to x} k(y)$.

Then $2M \geq l \geq k$ on X, and $l = k$ on $X - N'$, that is, almost everywhere for $\mu$, therefore $\int l \, d\mu \leq \varepsilon / 2$. On the other hand, $l$ is bounded and upper semi-continuous, hence is the lower envelope of the set of bounded continuous functions $\geq l$. Therefore there exists a bounded continuous function $h \geq l$ on X such that $h \leq 2M$ and $\int h \, d\mu \leq \int l \, d\mu + \varepsilon / 2$ (§ 4, No. 4, Cor. 2 of Prop. 5). Then $\int h \, d\mu \leq \varepsilon$ and $|f - g_1 a_1 - \cdots - g_n a_n| \leq h$.

(ii) $\Rightarrow$ (i). Suppose that condition (ii) is satisfied. For every $x \in X$ let $\omega(x)$ be the oscillation of f at x (GT, IX, § 2, No. 3). Let $\varepsilon > 0$. There exist $a_1, \ldots, a_n, g_1, \ldots, g_n, h$ with the properties of (ii). For every $x \in X$, $\omega(x)$ is the oscillation of $f - g_1 a_1 - \cdots - g_n a_n$ at x, therefore $\omega(x) \leq 2h(x)$. Thus $\int \omega \, d\mu \leq 2\varepsilon$. Consequently, the set $A_\varepsilon$ of $x \in X$ such that $\omega(x) \geq \sqrt{\varepsilon}$ satisfies $\mu(A_\varepsilon) \leq 2\sqrt{\varepsilon}$. This proves that $\mu(N) \leq 2\sqrt{\varepsilon}$, whence $\mu(N) = 0$.