Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$.

Question

Let $I$ be a generalized rectangle in $\Bbb R^n$

Suppose that the function $f\colon I\to \Bbb R$ is continuous. Assume that $f(x)\ge 0$, $\forall x \in I$

My idea is that

For $(\impliedby)$

Since $f\colon I\to \Bbb R$ is identically zero, $$f(I)=0$$

Then $$\int_{I}f=\int 0=0$$

For $(\implies)$,

Since $f$ is continuous, the function is integrable.

i.e $\int _{I} f $ exists.

I need the show that $\int f=0$ but how?

Hopefully, other solution is true. Please check this. And how to continue this? Thank you:)

You spelled it "identiacally". That was wrong. You did it three times, so it was most likely a systematic error rather than a random one. — dfeuer, Jul 27 '13 at 17:41
Oh! I didnt pay any attention. Sorry. I guess because of keybord dictionary. Thank you :) @dfeuer — Bstr, Jul 27 '13 at 17:49
I've posted a related question: http://math.stackexchange.com/questions/453580/might-proof-by-contradiction-be-needed-for-this-mundane-problem — Michael Hardy, Jul 27 '13 at 17:56

Michael Hardy · Answer 1 · 2013-07-27T17:21:53.117

This may be only a minor variation on an earlier answer, but maybe it adds something.

Suppose there's some point $x_0$ where $f(x_0)>0$. Let $\varepsilon=f(x_0)/2$. Then by continuity, there is some $\delta>0$ such that for $x$ in the open interval with endpoints $x_0\pm\delta$, the distance between $f(x)$ and $f(x_0)$ is less than $\varepsilon$. That means $f(x)>f(x_0)/2$ on that interval. Hence $$ \int_{x_0-\delta}^{x_0+\delta} f(x)\,dx > 2\delta\cdot\frac{f(x_0)}{2} = \delta f(x_0)>0. $$

Maybe one reason I feel I ought to post this is that there is a question: intuitively, the statement that if a function is positive on an interval, then its integral over that interval is postive, seems obvious. But how does one prove it without doing something like what I did above? What I did gives a partition of the interval, $\{a,x_0-\delta,x_0+\delta,b\}$, where $a,b$ are the endpoints, for which the lower Riemann sum is positive. Or if you like Lebesgue's definition, it gives a simple function dominated by $f$, whose integral is positive.

Pedro · Accepted Answer · 2013-07-27T16:31:19.387

3

One direction is obvious. If $f$ is identically $0$, the integral is zero. Now suppose $f$ is continuous but not identically zero. Then there exists a point in $I$ such that $f(\xi)>0$. By continuity there exists a neighborhood of $\xi$ where $f(x)>0$ whenever $x\in(\xi-\delta,\xi+\delta)$. Can you take it from here?

Since $f\geqslant 0$, it follows that $$\int_I f \geq \int_{(\xi-\delta,\xi+\delta)} f>0$$

Thus, we have proven the contrapositive.

edited Jul 27 '13 at 16:31

answered Jul 27 '13 at 15:51

Pedro

122,002

No. Please can you continue from here? Thank you. – Bstr Jul 27 '13 at 16:24
I want to learn incorrect. And I dont know how to reach the solution by using the definiton of continouty ? – Bstr Jul 27 '13 at 16:28
Okay Iunderstand thanks:) – Bstr Jul 27 '13 at 16:37
Hi, Dear @PeterTamaroff can you help me to solve this question as well? Thank you:) http://math.stackexchange.com/questions/453678/fundamental-optimization-question-consisting-of-two-parts – Bstr Jul 27 '13 at 22:06

Kunnysan · Answer 3 · 2013-07-27T17:47:09.883

3

Suppose $f(x)>0$ for some $x\in I$, as $f$ is continuous $\exists\ \delta>0$ such that $\forall y\in B_\delta(x)\cap I$, we will have $f(y)>0$. Then, as $f(x)\ge 0$

$$\displaystyle\int_If\ge\displaystyle\int_{B_\delta(x)\cap I}f>0 \implies\text{contradiction}$$.

edited Jul 27 '13 at 17:47

answered Jul 27 '13 at 15:56

Kunnysan

2,050

A downvote? Why? – Kunnysan Jul 27 '13 at 17:07
The downvote was probably because your answer is unclear. – dfeuer Jul 27 '13 at 17:43
@dfeuer I think it is clear, but I would choose another name for $x$. – Pedro Jul 27 '13 at 17:48

Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$.

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