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It seems straightforward, every time I draw a picture of a function that's always greater than zero, then the area under it is also always greater than zero. But then when I look at the definition of an integral, it's some convoluted summation that I don't use for anything. Is it actually true that if $f>0$ over some [a,b] then the $ \int_{a}^{b}fdx>0$? Because I can't find any function where that's not true, assuming there's no discontinuities.

user608672
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    The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities. – Matt Samuel Dec 05 '18 at 01:11
  • Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition. – user608672 Dec 05 '18 at 01:26
  • I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof. – Matt Samuel Dec 05 '18 at 01:42
  • Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct. – xbh Dec 05 '18 at 01:45
  • Riemann integral – user608672 Dec 05 '18 at 02:40

3 Answers3

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Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.

To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*\in[a,b]$.

$f(x_*)>0$ by hypothesis. Let $\epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $\delta>0$ such that $$\begin{align*}\left|x-x_*\right|<\delta\quad&\Rightarrow\quad\left|f(x)-f(x_*)\right|<\epsilon\\ &\Rightarrow\quad f(x)>f(x_*)-\epsilon=f(x_*)/2>0\text{.}\end{align*}$$

At this point, we have that within the non-empty interval $(\max(x_*-\delta,a),\min(x_*+\delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.

For any Riemann sum such that both $\max(x_*-\delta,a)$ and $\min(x_*+\delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(\min(x_*+\delta,b))-\max(x_*-\delta,a))f(x_*)/2>0$.

The integral is the limit as the Riemann sums are refined, so $$\int_a^bf(x)\,\mathrm{d}x\geq(\min(x_*+\delta,b))-\max(x_*-\delta,a))f(x_*)/2>0$$

obscurans
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A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.

We have that for any antiderivative of $f$, say $F$, that $\int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$

$$\frac{d}{dx}\int_a^x f(t)dt=F’(x)=f(x)>0.$$

So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $\int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $\int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.

John B
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Assume that $f$ is continuous.

By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.

Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+\frac{b-a}{N}]$, $[a+\frac{b-a}{N},a+2 \frac{b-a}{N}]$, ..., $[b-\frac{b-a}{N}, b]$ each of length $\frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$\int_a^b f(x)dx=\lim_{N \to \infty} \sum_{n=1}^N \frac{b-a}{N} f(a_n)\ge \lim_{N \to \infty} \sum_{n=1}^N \frac{b-a}{N}m=(b-a)m>0.$$

John B
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  • $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable. – obscurans Dec 05 '18 at 05:18
  • I will add the assumption that f is continuous – John B Dec 05 '18 at 05:34
  • But then you didn't prove the statement $f>0\Rightarrow\int_a^bf(x)dx>0$, you added conditions. – obscurans Dec 05 '18 at 05:43
  • True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.” – John B Dec 05 '18 at 05:49