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Measure and integration
Lesson 4 of 75
Challenge

Convergence theorems

Objective. Apply monotone and dominated convergence.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Convergence theorems describe when limits of integrals equal integrals of limits for sequences of measurable functions. The Monotone Convergence Theorem states that if 0≤f₁≤f₂≤... increases pointwise to f, then ∫fₙdμ→∫f dμ. The Dominated Convergence Theorem allows fₙ→f pointwise without monotonicity, provided |fₙ|≤g for some integrable g, giving ∫fₙdμ→∫f dμ. For example, fₙ(x)=x/n on [0,1] converges to 0 pointwise and is dominated by g=1, so ∫fₙ→0 by dominated convergence, matching direct calculation ∫₀¹(x/n)dx=1/(2n)→0.

Worked example

Using a convergence theorem, find lim_{n→∞} ∫₀¹ (nx)/(1+n²x²) dx.

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Illustrated card showing 1
  1. 1Let fₙ(x)=nx/(1+n²x²); note fₙ(x)→0 pointwise for each fixed x>0
  2. 2Bound fₙ(x) ≤ 1/2 for all x,n since nx/(1+n²x²)≤1/(2) by AM-GM
  3. 3Constant 1/2 is integrable on [0,1], so dominated convergence applies
  4. 4lim ∫₀¹fₙdx = ∫₀¹0 dx = 0

Answer: 0

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Common mistake. Students apply dominated convergence without finding an actual integrable dominating function; fₙ=n·𝟙_{(0,1/n)} converges pointwise to 0 but ∫fₙ=1 for all n because no integrable dominating function exists. Always exhibit g explicitly.

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