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:


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.

  • $\begingroup$ 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

$(i)$ 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.

  • $\begingroup$ 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

$\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 $[c,d] \subset [a,b]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.