From d5d03954dfafcc4ab7ea4a181d7fe5695608965b Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 16 Jun 2026 13:45:06 -0400 Subject: [PATCH] Slight typographic adjustment. --- src/measure/vector/fin.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/measure/vector/fin.tex b/src/measure/vector/fin.tex index c7eabd9..c9d7add 100644 --- a/src/measure/vector/fin.tex +++ b/src/measure/vector/fin.tex @@ -82,7 +82,7 @@ In any case, the above example shows that a linear functional on $M(X, \cm; \com Let $M = \sup\bracs{\norm{\nu_J}_{\text{var}}| J \subset I \text{ countable}}$, then there exists $J \subset I$ countable such that $\norm{\nu_J}_{\text{var}} = M$. For each $i \in I \setminus J$, \[ \norm{\nu_J}_{\text{var}} + \normn{\nu_a^{(i)}}_{\text{var}} - = \norm{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M + = \normn{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M \] By maximality of $M$, $\normn{\nu_a^{(i)}}_{\text{var}} = 0$, and by maximality of $\seqi{\mu}$, $\nu - \nu_J = 0$. Let $g \in [l^1(I); L^1(\mu_i; H)]$ be defined by $g_i = f_i$ for each $i \in I$, then $\nu = \nu_J = \mu_g$, and the mapping is surjective.