Let $f:[a,b]\rightarrow\mathcal R[a,b]$ (the set of Riemann integrable functions from $a$ to $b$, inclusive). If there exists an $M\in\mathbb{R}$ such that for all $x\in[a,b],|f(x)|\leq M$, show that

$$\left|\int_a^bf\right|\leq M(b-a)$$

By the definition of a Riemann integral function we know that there exists a real number $L=\int_a^bf$ such that for every $\varepsilon>0$ there exists $\delta>0$ such that if $\dot{\mathcal P}$ is any tagged partition of $[a,b]$ with $||\dot{\mathcal P}||<\delta$, then

$$|S(f;\dot{\mathcal P})-L|<\varepsilon$$

Where $S(f;\dot{\mathcal P})$ is the Riemann sum of $f$ corresponding to a tagged partition $\dot{\mathcal P}$.

Since $S(f;\dot{\mathcal P}) := \sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ and $|f(x)|\leq M, \forall x\in[a,b]$, we see that $S(f;\dot{\mathcal P})\leq\sum_{i=1}^nM(x_i-x_{i-1})=M\sum_{i=1}^n(x_i-x_{i-1})=M(b-a)$.

So we see that $S(f;\dot{\mathcal P})\leq M(b-a)$, but I am having trouble relating this to $\left|\int_a^bf\right|$. Am I on the right track?

  • $\begingroup$ See here. $\endgroup$ Commented Dec 14, 2015 at 6:38
  • $\begingroup$ @Mattos This is a good resource, but that doesn't say anything about my main issue that I could find, and that is that I do not know how to go from a statement about $S(f;\dot{\mathcal{P}})$ to a statement about $\int_a^bf$. $\endgroup$
    – flubsy
    Commented Dec 14, 2015 at 7:00
  • $\begingroup$ It's easier with upper and lower sums. $\endgroup$
    – egreg
    Commented Dec 14, 2015 at 13:00

2 Answers 2


You're almost there.

As $\vert S(f;\dot{\mathcal P}) -L \vert < \epsilon$, you also have $$L < S(f;\dot{\mathcal P}) + \epsilon \le M(b-a) + \epsilon$$

As this is true for all $\epsilon > 0$, you get $$L \le M(b-a)$$ Now you'll be done if you also prove $-M(b-a) \le L$. You'll get that from the inequalities $$\begin{cases} -\epsilon +S(f;\dot{\mathcal P}) < L\\ -M(b-a) \le S(f;\dot{\mathcal P}) \end{cases}$$


$|\int _a^bf|\leq \int_a^b |f|\leq M\int _a^b =M(b-a)$


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