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Pointwise Convergence of Functions
Suppose that X is a set, f1,f2,…:X→ℝ is a sequence of functions, and f:X→ℝ. We say that f1,f2,… converges pointwise on X to f if limk→∞⁡fk(x)=f(x) for every x∈X.
Uniform Convergence of Functions
Suppose that X is a set, f1,f2,…:X→ℝ is a sequence of functions, and f:X→ℝ. We say that f1,f2,… converges uniformly on X to f if for every ϵ∈ℝ+, there exists n∈ℤ+ such that |fk(x)−f(x)|<ϵ for every k≥n and every x∈X.
Uniform Limit of Continuous Functions is Continuous
Suppose that B⊆ℝ, f1,f2,…:B→ℝ converges uniformly on B to f:B→ℝ, and b∈B. If each fk is continuous at b, then f is continuous at b.
Egorov's Theorem
Suppose that (X,𝒮,μ) is a measure space with μ(X)<∞. If f1,f2,…:X→ℝ are 𝒮-measurable and converge pointwise on X to f:X→ℝ, then for every ϵ∈ℝ+, there exists E∈𝒮 such that μ(X∖E)<ϵ and f1,f2,… converges uniformly to f on E.
Simple Function
A function is simple if it takes only finitely many values.
Approximation by Simple Functions
Suppose that (X,𝒮) is a measurable space and f:X→[−∞,∞] is 𝒮-measurable. Then there exists a sequence f1,f2,…:X→ℝ such that
  • each fk is simple and 𝒮-measurable,
  • |fk(x)|≤|fk+1(x)|≤|f(x)| for every k∈ℤ+ and every x∈X,
  • limk→∞⁡fk(x)=f(x) for every x∈X,
  • if f is bounded, then f1,f2,… converges uniformly to f on X.
Luzin's Theorem
Suppose that g:ℝ→ℝ is Borel measurable. For every ϵ∈ℝ+, there exists a closed set F⊆ℝ such that |ℝ∖F|<ϵ and g|F is continuous on F.
Continuous Extensions from Closed Subsets of the Real Line
Suppose that F⊆ℝ is closed and g:F→ℝ is continuous. Then there exists a continuous function h:ℝ→ℝ such that h|F=g.
Luzin's Theorem, Extension Version
Suppose that E⊆ℝ and g:E→ℝ is Borel measurable. For every ϵ∈ℝ+, there exists a closed set F⊆E and a continuous function h:ℝ→ℝ such that |E∖F|<ϵandh|F=g|F.
Lebesgue Measurable Function
Suppose that A⊆ℝ. A function f:A→ℝ is Lebesgue measurable if f−1(B) is a Lebesgue measurable set for every Borel set B⊆ℝ.
Every Lebesgue Measurable Function is Almost Borel Measurable
Suppose that f:ℝ→ℝ is Lebesgue measurable. Then there exists a Borel measurable function g:ℝ→ℝ such that |{x∈ℝ:g(x)≠f(x)}|=0.