Integration in R^n

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 991

Here's a problem I am reviewing from analysis. I will list the problem with some commentary below:

Problem: Let $A \subset \mathbb{R}^n$ be open and have volume, and let $f: A \rightarrow \mathbb{R}^n$ be continuous, $f(x) \geq 0$ and f(x_0) > 0

for some $x_0 \in A$ .
Show that $$\int_A f > 0$ .

Ok, I think saying that A has volume (along with continuity) together forces the integral to exist. I don't think I need to say anything else(?) about this. Now, I remember seeing this back when I did the single variable integral, and I know that continuity means that there exists $\delta > 0$ such that f(x_*) >0

for all $x_* \in (x_0 - \delta, x_0 + \delta)$ .
If my memory is correct, you can then write some inequality, like $$\int_A f \geq \int^{x_0 + \delta}_{x_0 - \delta} \text{dx} = 2\delta > 0$

I think that is the Calculus II approach. How can I modify this for the $\mathbb{R}^n$ setting?

Shadow · **Posted:** Wed, 7 Mar 2012 19:41:13 UTC

Justin wrote:

Here's a problem I am reviewing from analysis. I will list the problem with some commentary below:

Problem: Let $A \subset \mathbb{R}^n$ be open and have volume, and let $f: A \rightarrow \mathbb{R}^n$ be continuous, $f(x) \geq 0$ and f(x_0) > 0

for some $x_0 \in A$ .
Show that $$\int_A f > 0$ .

Ok, I think saying that A has volume (along with continuity) together forces the integral to exist. I don't think I need to say anything else(?) about this. Now, I remember seeing this back when I did the single variable integral, and I know that continuity means that there exists $\delta > 0$ such that f(x_*) >0

for all $x_* \in (x_0 - \delta, x_0 + \delta)$ .
If my memory is correct, you can then write some inequality, like $$\int_A f \geq \int^{x_0 + \delta}_{x_0 - \delta} \text{dx} = 2\delta > 0$

I think that is the Calculus II approach. How can I modify this for the $\mathbb{R}^n$ setting?

Break up the integration domain into two pieces, one on which $f>\delta$ for some fixed delta and the other you just care that it's nonnegative. Add the two. In this case you just get $\delta\cdot V_{n,f}$ where $V_{n,f}$ is the size of the ball on which you have a greater than delta value for

.

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 991

Thanks, that definitely works.

Shadow · **Posted:** Wed, 7 Mar 2012 22:12:48 UTC

Justin wrote:

Thanks, that definitely works.

It does!? MADNESS! ;-)

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Integration in R^n

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