Lemma 3

— Suppose that the positive measure $\nu$ is such that every $\mu$-negligible compact set is $\nu$-negligible. Let $\mathfrak{K}$ be the set of compact subsets $K$ of $T$ having the following property:

(11) There exists a constant $M \geq 0$ such that $\varphi_K \cdot \nu \leq M \varphi_K \cdot \mu$. The set $\mathscr{K}$ is then $\mu$-dense in $T$.

If $K$ satisfies (11), and if $A$ is a Borel set contained in $K$, it follows at once from Prop. 8 that $\varphi_A \cdot \nu \leq M \varphi_A \cdot \mu$; from this, one deduces that the union of two elements $K, K'$ of $\mathscr{K}$ belongs to $\mathscr{K}$ because $\varphi_{K \cup K'} = \varphi_K + \varphi_A$, where $A = K' \cap \complement K$. To establish the lemma, it remains to prove that every compact set $L$ such that $\mu(L) > 0$ contains a compact set $K \in \mathscr{K}$ such that $\mu(K) > 0$ (Ch. IV, §5, No. 8, Prop. 12). Choose a number $M > \nu(L)/\mu(L)$ and apply Lemma 1 to the bounded positive measure $\alpha = \varphi_L \cdot (\nu + M \mu)$ and the measure $\beta = \varphi_L \cdot (\nu - M \mu)$. Replacing if necessary the function $u$ such that $\beta = u \cdot \alpha$ by a function equal to it $\alpha$-almost everywhere, one can suppose that $u$ is universally measurable (§3, No. 4, Prop. 7) and is zero outside of $L$. The set $H$ of $t \in T$ such that $u(t) < 0$, which is contained in $L$, could not be $\mu$-negligible, for it would then be $\nu$-negligible (by the remark made at the beginning of the proof of Th. 2), hence $\alpha$-negligible, and one would have $\beta(L) > 0$, which contradicts the choice of $M$. Let $K$ be a compact set contained in $H$, such that $\mu(K) > 0$; let us show that $K \in \mathscr{K}$, which will establish the lemma. By Prop. 8,

The function $\varphi_K u$ is negative, therefore we indeed have $\varphi_K \cdot \nu \leq M \varphi_K \cdot \mu$.

Let us now complete the proof of Theorem 2. Assume that the condition 3) is verified and define $\mathscr{K}$ as in Lemma 3. Let $(K_\alpha)_{\alpha \in A}$ be a locally countable family of pairwise disjoint elements of $\mathscr{K}$, such that the set $N = T - \bigcup_{\alpha \in A} K_\alpha$ is locally $\mu$-negligible (Ch. IV, §5, No. 9, Prop. 14); the family $(K_\alpha)$ being locally countable, $N$ is universally measurable and therefore locally $\nu$-negligible. Set $\mu_\alpha = \varphi_{K_\alpha} \cdot \mu$, $\nu_\alpha = \varphi_{K_\alpha} \cdot \nu$; since the functions $\varphi_{K_\alpha}$ form a locally countable family, whose sum is equal to 1 locally almost everywhere for $\mu$ and for $\nu$, Proposition 6 implies that $\mu = \sum_{\alpha \in A} \mu_\alpha$, $\nu = \sum_{\alpha \in A} \nu_\alpha$. On the other hand, by the definition of $\mathscr{K}$, there exists for every $\alpha$ a constant $M_\alpha$ such that $\nu_\alpha \leq M_\alpha \mu_\alpha$; Lemma 2 therefore implies the existence of a function $g_\alpha$, which one can suppose to be zero outside of $K_\alpha$ and positive (Cor. 3 of Prop. 3), such that $\nu_\alpha = g_\alpha \cdot \mu_\alpha$. Therefore (No. 4, Prop. 8)

Set $g = \sum_{\alpha \in A} g_\alpha$; since the family $(g_\alpha)$ is locally countable and the family $(\nu_\alpha)$ is summable, Proposition 6 implies that $g$ is locally $\mu$-integrable, and that $\nu = g \cdot \mu$, which establishes the theorem.