# 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.

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?

• GEdgar, Sorry - those were typos. – johnnyboy23 Apr 2 at 21:35

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.
• I'm confused about this step: $$\int_{X+} |g|\,\mathrm{d}\sigma + \int_{X-} |g|\,\mathrm{d}\sigma \\ = \int_{X_+} g\,\mathrm{d}\sigma - \int_{X_-} g\,\mathrm{d}x\\$$ – johnnyboy23 Apr 2 at 22:08
• @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. – rolandcyp Apr 2 at 22:10