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Proposition.

Let $g:X \to \overline{\mathbb{R}}$ be $\sigma$-integrable. If $|\int g \, d\sigma| = \int|g| d\, \sigma$, then $g \geq 0$ a.e. or $g \leq 0$ a.e.

Proof

Suppose that $|\int g \, d\sigma| = \int|g| d\, \sigma$ and $g \not\leq 0$ almost everywhere. Then we know that $\sigma([g> 0])$ has positive measure. We need to show that $g \geq 0$ almost everywhere, that is, $\sigma([g<0]) = 0$.

Any ideas on how to proceed?

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  • $\begingroup$ GEdgar, Sorry - those were typos. $\endgroup$ – johnnyboy23 Apr 2 at 21:35
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First, by replacing $g$ with $-g$, we may assume that $$ \int_X g\,\mathrm{d}\sigma \geq 0. $$ Then, we must have \begin{align*} \int_X g_+\,\mathrm{d}\sigma - \int_X g_-\,\mathrm{d}\sigma = \int_X g\,\mathrm{d}\sigma = \left\vert \int_X g\,\mathrm{d}\sigma \right\vert &= \int_X |g|\,\mathrm{d}\sigma\\ &= \int_X (g_+ + g_{-})\,\mathrm{d}\sigma\\ &= \int_X g_+\,\mathrm{d}\sigma + \int_X g_-\,\mathrm{d}\sigma. \end{align*} However, this would imply that $$ 0 = 2 \int_X g_-\,\mathrm{d}\sigma $$ whence it follows that $g_- = 0$ almost everywhere. From here, we can infer that $g \geq 0$ almost everywhere.

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  • $\begingroup$ I'm confused about this step: $$\int_{X+} |g|\,\mathrm{d}\sigma + \int_{X-} |g|\,\mathrm{d}\sigma \\ = \int_{X_+} g\,\mathrm{d}\sigma - \int_{X_-} g\,\mathrm{d}x\\$$ $\endgroup$ – johnnyboy23 Apr 2 at 22:08
  • $\begingroup$ @johnnyboy23 If $x \in X_{+}$ then $g(x) > 0$ and if $x \in X_-$ then $g(x) < 0$. Consequently, we have $|g(x)| = g(x)$ and $|g(x)| = -g(x)$ respectively. $\endgroup$ – rolandcyp Apr 2 at 22:10
  • $\begingroup$ Got it. Thanks. $\endgroup$ – johnnyboy23 Apr 2 at 22:10

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