I have that every Riemann sum $$S := |\sum_{i=1}^n f(t_i)(x_i - x_{i-1})| \le \sum_{i = 1}^n M(x_i - x_{i-1}) = M(b-a),$$ where $t_i$ is an arbitrary $x \in (x_{i-1}, x_i)$, and $x_1 \le ... \le x_n$ is a partitioning of $[a, b].$

I'm not sure how to extend that to $|\int_a^b f(x)| \le M(b-a).$

Does anyone have any pointers?


  • $\begingroup$ Hint: How are Riemann-sums and Riemann-integrals related? $\endgroup$ – sranthrop Nov 29 '15 at 23:42
  • $\begingroup$ @sranthrop I know that the Riemann sums are arbitrarily close to the integral, using the epsilon-delta definition, but I'm not sure how to formally state that if all of the sums are less than some constant C, than the integral is also less than C. $\endgroup$ – mxdg Nov 29 '15 at 23:50
  • $\begingroup$ One device to remember in most situations. If I wish to prove that $|\int_a^b f(x)\,dx| \leq M(b-a)$ I could do that by proving instead that, for any $\epsilon>0$, this inequality $$\left|\int_a^b f(x)\,dx\right | \leq M(b-a) + \epsilon$$ must be true. $\endgroup$ – B. S. Thomson Nov 29 '15 at 23:54
  • $\begingroup$ @B.S.Thomson I'm not sure how proving the second part shows the first part is true. I see how I could go from the statement without epsilon to the statement with epsilon. $\endgroup$ – mxdg Nov 30 '15 at 0:10
  • 1
    $\begingroup$ Then your life is a lot easier. The functions $g_1(x)=-M$ and $g_2(x)=M$ are integrable and so $$\int_a^b (-M)dx \leq \int_a^b f(x)\, dx \leq \int_a^b M \,dx$$ and you just have to remember how to integrate constant functions. $\endgroup$ – B. S. Thomson Nov 30 '15 at 2:07

For any partition $\mathcal{P} = \{a = t_0 < t_1 < \cdots < t_n = b\}$ of $[a,b]$ we know that $$ \int_a^b f \le U(f, \mathcal{P}) $$ where $$ U(f, \mathcal{P}) = \sum_{i=0}^n \sup_{[t_{i-1}, t_i]} f(x) (t_i - t_{i-1}) $$ is the upper Riemann sum of $f$ with the partition $\mathcal{P}$. Furthermore, as you state in your question, you know that $$ \left|U(f, \mathcal{P})\right| \le M(b-a). $$ Can you see how to combine these facts to obtain the result?

  • $\begingroup$ Is there a source/proof that I can take a look at for the first part? That's the part that I've been trying to find in my textbook but I haven't been able to find. $\endgroup$ – mxdg Nov 30 '15 at 0:42
  • 1
    $\begingroup$ If your textbook just uses an $\epsilon$, $\delta$ version of the integral then you would have to study this upper/lower version as well to use it. Alternatively carry on in the direction you were pursuing. For any $\epsilon>0$ you can find a partition with $$ \left| \int_a^b f - \sum_{i=1}^n f(t_i)(x_i - x_{i-1})\right| < \epsilon$$ and deduce that $$\left| \int_a^b f \right| <\left| \sum_{i=1}^n f(t_i)(x_i - x_{i-1})\right| + \epsilon.$$ $\endgroup$ – B. S. Thomson Nov 30 '15 at 0:47

Hint: If you think about it, this comes down to prove that $$\left|\int_a^b f(x) \mathrm{d}x\right|\leq \int_a^b |f(x)|\mathrm{d}x.$$

To help you, let us prove that $|f|:x\mapsto |f(x)|$ is integrable. Given a function $g:[a,b]\rightarrow \mathbb{R}$, we have that $g$ is integrable iff $\forall \varepsilon > 0$, there exists a partition $\mathcal{P}_n=\{x_0,...,x_n\}\in[a,b], x_0 = a, x_n=b,$ such that

$$\sum_{i=1}^n \omega_i(g)(x_i-x_{i-1})<\varepsilon,$$

where $\omega_i(g) = \sup g([x_{i-1},x_i]) - \inf g([x_{i-1},x_i])= \sup\{|g(x)-g(y)|,x,y\in[x_{i-1},x_i]\}$.

Since $$||f(x)|-|f(y)||\leq |f(x)-f(y)|,$$ we can conclude that for every partition $\mathcal{P}_n$, $\omega_i(|f|)\leq\omega_i(f)$. Thus, $|f|$ is integrable.

Furthermore, $-|f(x)|\leq f(x)\leq |f(x)|$, and therefore

\begin{align}&&-\int_a^b |f(x)|\mathrm{d}x\leq \int_a^b f(x)\mathrm{d}x\leq \int_a^b |f(x)|\mathrm{d}x&\\ \Leftrightarrow &&\left|\int_a^b f(x)\mathrm{d}x\right|\leq \int_a^b |f(x)|\mathrm{d}x.&\end{align}

  • $\begingroup$ "...which gives us the inequality we wanted." Normally yes, but maybe our student doesn't yet yet know how to prove the inequality property for integrals. I'm finding it hard to answer questions posed on this site since the students are often unable to explain what facts are available to them. $\endgroup$ – B. S. Thomson Nov 30 '15 at 1:12

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.