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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?

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  • $\begingroup$ 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. $\endgroup$ – Fnacool Mar 12 '16 at 1:35
  • $\begingroup$ check out "Proposition 9" on page 80 in this. what is that named? $\endgroup$ – robert bristow-johnson Mar 12 '16 at 1:44
  • $\begingroup$ @robertbristow-johnson, page 80 in that pdf is blank. $\endgroup$ – lhf Mar 12 '16 at 1:53
  • $\begingroup$ 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. $\endgroup$ – RRL Mar 12 '16 at 2:31
  • $\begingroup$ @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. $\endgroup$ – robert bristow-johnson Mar 12 '16 at 2:46
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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 $$\lVert f - g\rVert = \int_a^b |f-g|\,dx$$

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  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Mar 12 '16 at 2:51
  • $\begingroup$ @robert bristow-johnson The Cantor set is the set where the Cantor function does not have zero derivative. $\endgroup$ – Henricus V. Mar 12 '16 at 2:52
  • $\begingroup$ but Henry, that's a countable set. ain't it? $\endgroup$ – robert bristow-johnson Mar 12 '16 at 2:53
  • $\begingroup$ @robert bristow-johnson No. A bijection exists between the Cantor set and $[0,1]$. $\endgroup$ – Henricus V. Mar 12 '16 at 2:54
  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Mar 12 '16 at 2:55

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