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.
Arbejdstxrd eksempel · integrate e^(-x) dx from 0 to oo
Trin for trin
- \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.
Afslør svaret
Nu er du Vælg et problem, eller skriv eller tegne din egen. Hvert skridt, et billede, svaret skjult, indtil du spørger.
En gratis konto tilføjer noter om hver lektion, en rekord over, hvad du har afsluttet, dine løste problemer på ét sted, og en vejleder, du kan spørge om denne side. Matematik selv er åben for alle, logget på eller ej.
Tilmeld dig LoginSymboler brugt her
Tryk på ethvert symbol for den fulde definition, et billede, og hvad hvert bogstav i det betyder.
Hvordan: 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.
Spørgsmål folk spørger
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.