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Suppose $f$ is Riemann-integrable on $[a,b]$ such that $f(x)>0, \forall x \in [a,b].$

Prove: $$\int_a^bf(x)\mathop{dx}>0$$

Provided solution:

Suppose by contradiction that $I=\int_a^bf(x)\mathop{dx}=0.$

Let us take a sequence of normal partitions $T_n$ of $[a,b],$ that is partitons of $n$ intervals of length $\frac{b-a}{n}.$

Then for $n$ large enough, upper Darboux sum of $f$ is small as we wish.

So there exists $n_0$ such that for all $n>n_0:$

$$\tag{1}\sum_{i=0}^{n-1}\sup_{[x_i,x_{i+1}]}f\cdot\Delta x_i=\frac{b-a}{n}\sum_{i=0}^{n-1}\sup_{[x_i,x_{i+1}]}f<\frac{b-a}{2}$$

Therefore, there exists an interval $I_1:=[x_i,x_{i+1}],$ such that:

$$\tag{2}\sup_{[x_i,x_{i+1}]}f<\frac{1}{2}$$

By the integral monotonicity and positiveness of $f:$

$$\tag{3}0=\int_a^bf(x)\mathop{dx}\geq \int_{I_1}f(x)\mathop{dx} \geq 0$$

Hence, $\int_{I_1}f(x)\mathop{dx}=0.$

Repeating the process on $I_1,$ let us take a sequence of normal partitions of $[x_i,x_{i+1},$ that is intervals $[y_i,y_{i+1}]$ of length $\frac{x_{i+1}-x_i}{n}.$ Therefore, there exists $n_1$ such that for all $n>n_1:$

$$\tag{4}\sum_{i=0}^{n-1}\sup_{[y_i,y_{i+1}]}f\cdot\Delta x_i=\frac{x_{i+1}-x_i}{n}\sum_{i=0}^{n-1}\sup_{[y_i,y_{i+1}]}f<\frac{x_{i+1}-x_i}{4}$$

So there exists an interval $I_2:=[y_i,y_{i+1}] \subset I_1,$ such that:

$$\tag{5} \sup_{I_2}f<\frac{1}{4}$$

Continuing like that, we get a sequence of intervals: $\ \dots \subseteq I_2 \subseteq I_1,$ such that:

$$\tag{6} \sup_{I_n}f \leq \frac{1}{2^n}$$

By Cantor's intersection theorem:

$$\tag{7} \bigcap_{n=0}^\infty I_n \neq \emptyset $$

But, if $x \in \bigcap_{n=0}^\infty I_n,$ then $f(x) \leq \frac{1}{2^n},$ for all $n$, hence $f(x)=0,$ contradicting $f(x)>0.$


My questions:

$(a)$ At $(1),$ I know $\sum_{i=0}^{n-1} \Delta x_i = b-a,$ but why does it equal $\frac{b-a}{n}?$ And why is the inequality true?

$(a)$ At $(3),$ how is the monotonicity is used?

Any help is appreciated.

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  • $\begingroup$ Wouldn't the contradiction be $I \le 0$? $\endgroup$ – GFauxPas Jul 31 '17 at 19:20
  • $\begingroup$ Nice proof. Another approach is show that Riemann integrable functions are continuous somewhere. See this answer math.stackexchange.com/a/519921/72031 $\endgroup$ – Paramanand Singh Jul 31 '17 at 19:46
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$(i)$ The partition is normal (i.e., each subinterval is of equal length), hence $\Delta x_i = \frac{b-a}{n}$. This was simply factored out of the sum in step $(1)$.

$(ii)$ Monotonicity is used to show that $\int_{[a,b]}f \geq \int_{I_1} f$. This is because $\int_{[a,b] \setminus I_1} f \geq 0$ since $ f \geq 0$.

This is a neat proof, by the way. First one I've seen w/o using Lebesgue's characterization.

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  • $\begingroup$ Agree with your last line. It's the first proof for me also without using any sophisticated tools. +1 $\endgroup$ – Paramanand Singh Jul 31 '17 at 19:50
  • $\begingroup$ I see it now. Could you explain the inequality in $(1)?$ $\endgroup$ – Itay4 Aug 1 '17 at 5:44
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$\Delta x_i=\frac {b-a}n $ because b-a is divided into n equal subintervals.

Integration is monotonic in the sense that $f \ge0 \implies \int_a^b f \ge \int_c^d f$ when $[c,d] \subset [a,b]$.

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