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Suppose $f$ is integrable on [$a$,$b$] and $f(x)>0$ for all $x$ in [$a$,$b$]. Prove that $\displaystyle\int_a^b f(x) \mathrm{d}x>0$.

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marked as duplicate by David Mitra, Thomas, rschwieb, Micah, Tara B Mar 7 '13 at 20:41

This question was marked as an exact duplicate of an existing question.

Show that a lower sum has positive value. – Tyler Mar 7 '13 at 19:57
Well,I've been trying to do so,but haven't succeeded yet.In fact,that's what I'm asking here. – user1978522 Mar 7 '13 at 20:06
Well you didn't give any information other than the problem statement. Perhaps you should, so we would know what you've tried. – Tyler Mar 7 '13 at 20:09

For any Riemann sum on $\,[a,b]\,$ with partition $\,\{a=x_0<x_2,\ldots <x_n=b\}\,$ , we get

$$\sum_{n=0,\,||\Delta_x||\to 0}f(c_i)(x_i-x_{i-1})>0\ldots$$



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You've just written what I'm asking to prove. – user1978522 Mar 7 '13 at 20:08
I don't understand then: what isn't clear here? Both factors of each summand in that sum are positive! – DonAntonio Mar 7 '13 at 20:09
I think the hard part of the problem is to show strict inequality. How do you deduce that the limit of your Riemann sums, as the norms tend to zero, is positive from the fact that any particular sum is positive? – David Mitra Mar 7 '13 at 20:10
I think you're right, @DavidMitra...but I'd rather wait until the OP explains where he is stuck and/orwhat's he done so far. – DonAntonio Mar 7 '13 at 20:13

If $f$ is Riemann integrable on $[a,b]$ then the set of points $x$ at which $f$ is not continuous has measure zero. So we may choose some $x_1$ with $a<x_1<b$ for which $f$ is continuous at $x_1$. Since by assumption $k=f(x_1)>0$ we may choose some small $\delta$ so that $f(x) \ge k/2$ for $x_1-\delta \le x \le x_1+\delta$. Then since $f(x)>0$ on $[a,b]$ we have $\int_a^b f(x)dx \ge (2 \delta)\cdot k/2$. Here it is understood that $\delta$ is small enough that the interval $[x_1-\delta,x_1+\delta]$ is contained in $[a.b]$.

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$$M_i=\ supf(x) , x\in[x_{i-1},x_i]$$ $U(p,f)=$$\sum_{i=1} ^n $$M_i$ $\Delta_{x_i}$ such that [$x_i,x_{i+1}$] $i=0,1,2,...,n-1$ is arbitrary partition in [a,b] $$\displaystyle\int_a^b f(x) \mathrm{d}x=inf_pU(p,f) ,\forall p$$ $$ f\gt0 \to supf \gt0 \to M_i\gt0$$ then $U(p,f)\gt0$ Finlay conclude $$\displaystyle\int_a^b f(x) \mathrm{d}x>0$$

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The $>$ should be a $\ge$. – copper.hat Mar 7 '13 at 19:53

Ok, so I hope that you know that (with obvious meaning of notation) (all depending on your definition of the integral) $$ \int_a^b f(x)\; dx \geq \sum_{i}[\inf_{[x_i, x_{i+1}]} f(x)]\Delta x $$ for all partitions of $[a,b]$, where $\inf$ is the infimum of $f$ over each subinterval.

Claim: There is a $\delta >0$ and a subinterval $[x_1 , x_{i+1}]$ such that $f(x) \geq \delta$ for $x\in [x_i, x_{i+1}]$.

Proof Suppose this wasn't true. Then $$ \sum_{i}[\inf_{[x_i, x_{i+1}]} f(x)]\Delta x = 0 $$ for all pertitions. And so (taking the limit) the integral would be $0$. $\quad\quad\quad\square$

So there is an interval $[x_i, x_{i+1}]$ and a $\delta > 0$ such that $f(x)\geq \delta$ for all $x\in [x_i, x_{i+1}]$.

So now you have that $f(x)\geq \delta >0$ on an interval $[x_i , x_{i+1}]$. So then $$ \int_{x_i}^{x_{i+1}} f(x)\; dx \geq \delta(x_{i+1} - x_i) > 0. $$ So $$ \int_{a}^b f(x)\; dx = \int_a^{x_i} f(x)\; dx + \int_{x_i}^{x_{i+1}} f(x)\; dx + \int_{x_{i+1}}^b f(x)\; dx > 0. $$ (Assuming that you are fine with saying that the first and last integral definitely are greater then or equal to $0$).

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I don't see any contradiction in the integral being 0 . – user1978522 Mar 7 '13 at 20:25
@user1978522: Ok, let me add the last bit. – Thomas Mar 7 '13 at 20:26
It seems that you misunderstood me.What I was saying is that when you are taking the limit of that sum there is no contradiction,because from f(x)>0 it follows that the integral is either 0 or greater than zero ,so you should prove that it can't be zero ,otherwise you are using for your proof what is needed to prove. – user1978522 Mar 7 '13 at 20:38
@user1978522: Do you agree that if $f(x) \geq \delta$ on some subinterval, then the integral is greater than 0$? I answered the other question that this was closed as a duplicate of. – Thomas Mar 7 '13 at 20:43
Your argument has a flaw before that part so it's pointless to even consider it. Your argument is wrong in this part : Then $$ \sum_{i}[\inf_{[x_i, x_{i+1}]} f(x)]\Delta x = 0 $$ for all pertitions. And so (taking the limit) the integral would be $0$. Here you assume a contradiction and build the rest of your argument but what i'm saying is that there is no contradiction – user1978522 Mar 7 '13 at 20:51

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