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May 9th, 2017, 05:19 AM   #1
Glo
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$a,b\geq 0$ 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
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May 19th, 2017, 03:38 AM   #2
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Specifically, and we are told that the right side is finite.
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May 19th, 2017, 06:10 AM   #3
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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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