# What is the name of this theorem or lemma?

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?

• 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 1:35
• check out "Proposition 9" on page 80 in this. what is that named? – robert bristow-johnson Mar 12 '16 at 1:44
• @robertbristow-johnson, page 80 in that pdf is blank. – lhf Mar 12 '16 at 1:53
• 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 2:31
• @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 2:46

$f = g$ 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 $$\lVert f - g\rVert = \int_a^b |f-g|\,dx$$
• 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 2:51
• @robert bristow-johnson No. A bijection exists between the Cantor set and $[0,1]$. – Henricus V. Mar 12 '16 at 2:54
• 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 2:55