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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.

Àwọn ààtò àwọn ìṣàmúlò-ètò · integrate e^(-x) dx from 0 to oo

Calculus · Definite-Integrals

Integrate e^(-x) from 0 to oo

\int_{0}^{\infty} e^{- x}\, dx

Àwọn ìṣàmúlò-ètò

\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.

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Bii o ṣe le: 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.

Àwọn Àtòjọ-ẹ̀yàn

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.

Diẹ̀ nínú Measure Theory

σ-algebras and measures