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I'm studying Riemann Integral on measure theory class.

There is a function $f : [a,b] \to \mathbb{R}$ that is increasing or decreasing.

Is that function f is Riemann integrable? And if then, what are appropriate step functions?

And is that function continuous a.e.?

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2 Answers 2

up vote 3 down vote accepted

Yes, a monotonic function on an interval is Riemann integrable. In fact, one can come up with the step functions naively.

Let $P_n$ denote the partition $\{a, a + \frac{b-a}{n}, a + 2\frac{b-a}{n}, \dots, b\} =: \{x_0, \dots, x_{n}\}$. Then it's easy to see that $x_k - x_{k-1} = \frac{b-a}{n}$. Suppose wlog that $f$ is monotonic decreasing (increasing is more or less the same, or do the same reasoning on $-f$). Then if $M_k$ and $m_k$ denote the maximum and minimum on the $k$th part of the partition, then $M_k = f(x_{k-1})$ and $m_k = f(x_k)$.

Thus
1. (lower sum) $\displaystyle L(f,P_n) = \sum m_k(x_k - x_{k-1}) = \sum f(x_k)\frac{b-a}{n}$ and
2. (upper sum) $\displaystyle U(f,P_n) = \sum M_k(x_k - x_{k-1}) = \sum f(x_{k-1})\frac{b-a}{n}$

So then $$\begin{align} U(f,P_n) - L(f,P_n) &= \frac{b-a}{n}\left( \sum_{k = 1}^n f(x_{k-1}) - f(x_k)\right) = \\ &= \frac{b-a}{n} (f(x_0) - f(x_n)) = \frac{b-a}{n}(f(a) - f(b)) \\ &\to 0 \text{ as } (n \to \infty) \end{align}$$

It's not so hard to show that refinements of these partitions do not change the result.

You also ask whether a monotonic function on $[a,b]$ is continuous a.e. It is, in the sense that there are only a countable number of discontinuities. In particular, you might know that a sum of positive numbers is finite only if it has a countable number of nonzero terms (an easy way to see this - if there are uncountably many, then you can show there are uncountably many rationals as follows: consider the intervals implied by the sum, where each summand gives another interval. Each interval contains a rational.) The function is bounded on $[a,b]$ and monotonic, and so the sum of the 'jumps' at the points of discontinuity is finite. Thus there are at most countably many points of discontinuity.

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It's somehow complicated but concrete proof. Thank you @mixedmath –  wowhapjs Jun 10 '12 at 13:21

Yes it's Riemann integrable. The set of points of discontinuity of this function is at most countable and this function belongs to a larger class of functions which is class of functions who have right and left limit at every point inside $[a,b]$ and left limite at $b$ and right limit at $a$. Each function in the class is an uniform limit of sequence of step functions.

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Suppose for example that $f$ is increasing. To proof that $f$ has a left limit at $x_0 \in ]a,b[$, you can put : $\ell= \sup (f([a,x_0[)$ then show that $\displaystyle \ell=\lim_{x \to x_0, x< x_0} f(x) $ by using definition of $\sup$. –  Mohamed Jun 10 '12 at 12:16
    
Ah appropriate example and explain. Thank you @Mohamed –  wowhapjs Jun 10 '12 at 12:46

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