On that convolution race with Henry.
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\item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
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\item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
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\end{enumerate}
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\end{enumerate}
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Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
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Conversely, if $F: \real \to \real$ is a Stieltjes function, then
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\begin{enumerate}
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\item[(1C)] There exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ such that for all $-\infty < a < b < \infty$,
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\[
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\mu_F((a, b]) = F(b) - F(a)
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\]
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\end{enumerate}
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In which case, $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
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\end{definition}
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\end{definition}
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\begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
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\begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
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(1): Let
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(1): Let
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@@ -71,7 +81,7 @@
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(U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
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(U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
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Converse: Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
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(1C): Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
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\[
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\[
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\alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
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\alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
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\]
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\]
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