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Prove that $\int_a^b f(x)\,dx \gt 0$ if $f \geq 0$ for all $x \in [a,b]$ and $f$ is continuous at $x_0 \in [a,b]$ and $f(x_0) \gt 0$

EDIT. Please ignore below. It is very confusing actually -.-


Note: After typing this all out, I think I realized that my proof is complete, but I'll just post it to make sure :)

Attempt:

Find a partition $P = \{t_0,\ldots,t_n\}$ of $[a,b]$ s.t. $f(x)\gt f(x_0)/2$ for any $x \in [t_{i-1},t_i]$

Thus, the lower sum, $L(f,P)\geq x_0 (t_{i}-t_{i-1})/2 > 0$

[Basic Idea: since f is continuous at $x_0$, in the worst case scenario (i.e. $f=0$ at all points except in a nbhd of $x_0$) there must be some "bump" at $x_0$, which prevents the integral from actually equaling $0$. If we can find just one lower sum of a partition to be $> 0$, we will be done (I think...)]

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[I am not sure if your attempt is correct or not, but here's an alternate approach.] Instead of resorting to Riemann sums, perhaps you can use the monotonicity property. By continuity, there exists an open neighborhood $(x_0 - r, x_0 + r)$ where $f$ is at least $f(x_0)/2$. Now, consider the function $g = \frac{f(x_0)}{2} \mathbf 1_{(x_0 - r, x_0 + r)}$. Observe that $f \geqslant g$ and the integral of $g$ is positive. So by monotonicity of integration, the integral of $f$ is positive as well. –  Srivatsan Nov 16 '11 at 22:10
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@Srivatsan Thanks. I think my attempt was supposed to be something like that, but I see it's really messy –  MathMathCookie Nov 17 '11 at 23:01
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1 Answer 1

up vote 2 down vote accepted

I take it for granted that you know that this integral is never negative. So you need to give a single interval over which it's positive.

$f(x_0) > 0, f(x)$ continuous there means that there is a $\delta$ such that for $|x-x_0|< \delta$, $|f(x) - f(x_0)| < \frac{f(x_0)}{2}$.

So consider the interval $[x-\delta, x+\delta]$.

I think this is the easiest way to do it, perhaps.

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