Slight typographic adjustment.

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.