My Math Forum Integrability of the sum of integrable functions

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 May 9th, 2017, 05:19 AM #1 Newbie   Joined: Dec 2014 From: Italy Posts: 5 Thanks: 0 $a,b\geq 0$ and $a<+\infty$, $b<+\infty$, then $a-b<+\infty$? Let be $f\in C^0(\mathbb{R})$, such that $f\geq 0$. I suppose that $\int_{\mathbb{R}}f(t) dt<+\infty$. Under these hypothesis, can I say that $\int_{0}^{+\infty} f(t) dt-\int_{-\infty}^{0} f(t)dt<+\infty$ ? I think yes, since $\int_{\mathbb{R}}f(t) dt<+\infty$ iff $\int_{0}^{\infty}f(t)dt<+\infty$ e $\int_{-\infty}^{0}f(t)dt<+\infty$. If $a,b\geq 0$ and $a<+\infty$, $b<+\infty$, then $a-b<+\infty$. The viceversa of this last statement is not true, right? Am I right? Thanks! Last edited by Glo; May 9th, 2017 at 05:47 AM. Reason: more explicative title
 May 19th, 2017, 03:38 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,104 Thanks: 850 Specifically, $\left|\int_0^{\infty} f(t)dt- \int_{-\infty}^0 f(t)dt\right|\le \left|\int_{-\infty}^\infty f(t) dt\right|$ and we are told that the right side is finite.
 May 19th, 2017, 06:10 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,274 Thanks: 2435 Math Focus: Mainly analysis and algebra Suppose that $\displaystyle \int_0^{+\infty} f(t)\,\mathrm dt = 1$ and $\displaystyle \int_{-\infty}^0 f(t)\,\mathrm dt = -1$ then \begin{align*} \left| \int_0^{+\infty} f(t)\,\mathrm dt -\int_{-\infty}^0 f(t)\,\mathrm dt \right| &= |1 + 1| = 2 \\ \left| \int_{-\infty}^{+\infty} f(t)\,\mathrm dt \right| &= 0 \end{align*}

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