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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.

Eżempju maħdum · integrate e^(-x) dx from 0 to oo

Integrate e^(-x) from 0 to oo

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

Pass b'pass

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

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. u = - x,\quad du = -1\, dx

    Substitute u = - x.

  3. \int e^{- x}\, dx = \int - e^{u}\, d_u

    Rewrite the integral in terms of u.

  4. \int - e^{u}\, d_u = -1 \int e^{u}\, d_u

    Pull the constant -1 out of the integral.

  5. \int e^{u}\, d_u = e^{u}

    ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).

  6. = - e^{- x}

    Substitute back u = - x.

  7. F(\infty) - F(0) = \left(0\right) - \left(-1\right)

    Fundamental theorem of calculus: plug in the limits.

  8. = 1

    Simplify.

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Kif għandek tuża: The Lebesgue integral and convergence theorems

  1. First find an antiderivative F, then evaluate F(b) − F(a).
  2. Substitute u = - x.
  3. Rewrite the integral in terms of u.
  4. Pull the constant -1 out of the integral.
  5. ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
  6. Substitute back u = - x.
  7. Fundamental theorem of calculus: plug in the limits.
  8. Simplify.

Mistoqsijiet li n-nies jistaqsu

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.

Aktar fil Measure Theory