/sci/ - Science & Math

>>11593018
Any interval [math][a_{i-1}, a_i][/math] with [math]a_{i-1}<a_i[/math] contains both a rational and an irrational number. Then, the infimum of [math]f[/math] on the interval is at least smaller than [math]-1[/math] (which implies it is in fact -1), and the supremum is at least larger than [math]1[/math] (and thus is 1, yada yada).
>>11594581
Assume [math]f[/math] and [math]g[/math] are Riemann integrable, and their difference is a.e. zero. Then [math]\int _a ^b f(x) ~ dx - \int _a ^b g(x) dx = \int _a ^b [f(x)-g(x)] ~ dx[/math]. Since [math]f-g[/math] is a.e. zero, it's Lebesgue integrable even if [math]f[/math] and [math]g[/math] aren't ( source for the sake of it: http://mathonline.wikidot.com/lebesgue-integrability-of-functions-equalling-0-a-e-on-gener ), so we can use [math]\int_a ^b [f(x)-g(x)] ~ dx - \int _{[a, b]} [f(x)-g(x)] d \mu = 0[/math], and then finally [math]\int _a ^b f(x) ~ dx = \int _a ^b g(x) dx[/math].
I'm not sure if I did anything illegal or not, but I think it works as is.