maths.free › Measure Theory › The Lebesgue integral and convergence theorems
The Lebesgue integral and convergence theorems
Integrating by slicing the range instead of the domain; monotone and dominated convergence.
Riemann slices the x-axis; Lebesgue slices the y-axis and asks how much of the domain sits at each height. The payoff is the convergence theorems: under mild conditions, the limit of the integrals is the integral of the limit. Picture it: horizontal strips under the curve instead of vertical ones. Think it: L¹ and L² become complete spaces, the Hilbert spaces of functional analysis.
Үйлдэл · integrate e^(-x) dx from 0 to oo
Алхам алхмаар
- \int_{0}^{\infty} e^{- x}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- u = - x,\quad du = -1\, dx
Substitute u = - x.
- \int e^{- x}\, dx = \int - e^{u}\, d_u
Rewrite the integral in terms of u.
- \int - e^{u}\, d_u = -1 \int e^{u}\, d_u
Pull the constant -1 out of the integral.
- \int e^{u}\, d_u = e^{u}
∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- = - e^{- x}
Substitute back u = - x.
- F(\infty) - F(0) = \left(0\right) - \left(-1\right)
Fundamental theorem of calculus: plug in the limits.
- = 1
Simplify.
Хариулт
Одоо та Даалгавар сонго, эсвэл бич, эсвэл зур. Бүх алхам, зураг, хариулт нь асуулт асуух хүртэл нуугдаж байдаг.
Үнэгүй бүртгэл нь хичээл бүрийн тэмдэглэл, төгссөн хичээл, шийдсэн даалгаврын тэмдэглэл, энэ хуудсан дээр асууж болох багшийг нэгтгэж өгдөг. Математик нь бүртгэлтэй, бүртгэлтэйгүй хэн бүхэнд нээлттэй.
Бүртгүүлэх НэвтрэхЭнд ашигласан тэмдэгтүүд
Бүх тодорхойлолт, зураг, үсгийн утгыг харахын тулд ямар ч тэмдэгтийг дарна уу.
Яаж: The Lebesgue integral and convergence theorems
- First find an antiderivative F, then evaluate F(b) − F(a).
- Substitute u = - x.
- Rewrite the integral in terms of u.
- Pull the constant -1 out of the integral.
- ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- Substitute back u = - x.
- Fundamental theorem of calculus: plug in the limits.
- Simplify.
Хүмүүс асуудаг асуултууд
What is wrong with the Riemann integral?
It fails on functions that oscillate too much, and it does not interact well with limits: the limit of integrable functions need not be integrable. Lebesgue's integral fixes both.