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Real analysis: $$x, f(x), g(x) \in \mathbb{R}$$

If $f(x) = g(x)$ "almost everywhere" in the interval $a \le x \le b$ (that is every value in the interval other than no more than a countably infinite number of discrete "points" or values of $x$), then

$$ \int\limits_a^b f(x) \ dx = \int\limits_a^b g(x) \ dx $$

i remember this from Real Analysis in college (had a text by Royden or someone like that). what is this fact called?

robert bristow-johnson
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  • Not sure has a name. Its a corollary to monotonicity of the integral: if f is greater or equal to g, then the integral of f is greater or equal to integral of g. This (assuming linearity of the integral) is equivalent to the positivity of the integral: an integral of a nonnegative function is nonnegative. – Fnacool Mar 12 '16 at 01:35
  • check out "Proposition 9" on page 80 in this. what is that named? – robert bristow-johnson Mar 12 '16 at 01:44
  • @robertbristow-johnson, page 80 in that pdf is blank. – lhf Mar 12 '16 at 01:53
  • Riemann-Lebesgue Theorem enters here: A bounded function $f:[a,b] \to \mathbb{R}$ is Riemann integrable if and only if it is continuous almost everywhere. That $f = g$ a.e. implies $\int f = \int g$ is direct with Lebesgue integrals since the contribution from integrating over a measure zero set is zero. For Riemann integrals it also holds and can be proved by constructing partitions that enclose the discontinuity points in subintervals of arbitrarily small total length. – RRL Mar 12 '16 at 02:31
  • @lhf, page 80 as marked on the top of the pages. i think it's the 91th page of the pdf if you count the title page and TOC. – robert bristow-johnson Mar 12 '16 at 02:46

1 Answers1

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$f = g$ a.e. means that the set $\{x\mid f(x) \neq g(x)\}$ has $0$ Lebesgue measure. It might not be countable. Consider $f = 0$ and $g = \chi_S$, where $S$ is the Cantor set.

This statement is the identity of indiscernibles for the $L^1$ metric. In the $L^1$ space, (or the space of integrable functions on $[a,b]$) functions that agree a.e. are considered the same. This manifestation makes $L^1$ a metric space with the metric $$\lVert f - g\rVert = \int_a^b |f-g|\,dx$$

Henricus V.
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  • i'm gonna take your word for it, Henry. thank you. if no one else offers a competing answer, i'll likely check mark it as "answered". i remember the Cantor function (that is continuous, has derivative zero a.e., yet rises from 0 to 1 in the interval $0 \le x \le 1$) but i dunno what the Cantor set is. guess i'll have to look it up. – robert bristow-johnson Mar 12 '16 at 02:51
  • @robert bristow-johnson The Cantor set is the set where the Cantor function does not have zero derivative. – Henricus V. Mar 12 '16 at 02:52
  • but Henry, that's a countable set. ain't it? – robert bristow-johnson Mar 12 '16 at 02:53
  • @robert bristow-johnson No. A bijection exists between the Cantor set and $[0,1]$. – Henricus V. Mar 12 '16 at 02:54
  • hmmm. i thought that the set of all rational $x$ for the interval $0 \le x \le 1$ is a countable set. in fact, i remember how you order the set and map it to the positive integers. – robert bristow-johnson Mar 12 '16 at 02:55
  • @robert bristow-johnson Yes. $\mathbb{Q}$ is countable with zero measure. The Cantor set is essentially all numbers in $[0,1]$ whose ternary expansion has no $1$'s. – Henricus V. Mar 12 '16 at 02:57
  • but aren't they all rational??? like $\frac13$ $\frac23$ $\frac19$ $\frac29$...? and ain't a subset of a countable set also a countable set? – robert bristow-johnson Mar 12 '16 at 02:59
  • @robert bristow-johnson Try Cantor's diagonal argument. – Henricus V. Mar 12 '16 at 03:06
  • okay that argument is what i have seen to show that that the reals are not countable. one thing that it requires (if the numbers are binary) is that something like $0.bbbb01111111111....$ must be re-expressed as $0.bbbb10000000000...$ because you have two different bit patterns (or digit patterns) that converge to the same value. can you please just tell me what number in the Cantor set is irrational? if they are all rational, the set must be countable. – robert bristow-johnson Mar 12 '16 at 03:14
  • Irrationals in the Cantor set – Henricus V. Mar 12 '16 at 03:17
  • okay, i will mull this. this is beginning to get too hard. i've always thought that, what you call the Cantor set, was a subset of the rationals. in fact, i thought that i could rattle off an enumeration procedure that would cover the whole set: $\frac13, \frac23. \frac19, \frac29, \frac79, \frac89, \frac{1}{27}, \frac{2}{27}, \frac{7}{27}, \frac{8}{27}, \frac{13}{27}, \frac{14}{27}...$ it just seems to me that the set must contain only rational numbers. – robert bristow-johnson Mar 12 '16 at 03:21
  • Cantor's diagonal argument always work if you are unsure. Your enumeration never hits any non-repeating decimals. – Henricus V. Mar 12 '16 at 03:22