#lebesgue — Public Fediverse posts

Cédric #Villani, #Fields #Medal 2010 , On #Measure #Theory and #Lebesgue #Integration

https://youtube.com/watch?v=qBbq4RepjxE&si=nKL5OOYMsXEgIO1W

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes #Lebesgue integration more powerful than #Riemann integration. The theorem an be stated as follows:

Let \((f_n)\) be a sequence of measurable functions on a measure space \((\mathcal{S},\Sigma,\mu)\). Suppose that \((f_n)\) converges pointwise to a function \(f\) and is dominated by some Lebesgue integrable function \(g\), i.e. \(|f_n(x)|\leq g(x)\ \forall n\) and \(\forall x\in\mathcal{S}\). Then, \(f\) is Lebesgue integrable, and

\[\displaystyle\lim_{n\to\infty}\int_\mathcal{S}f_n\ \mathrm{d}\mu=\int_\mathcal{S}f\ \mathrm{d}\mu\]
#ConvergenceTheorem #Convergence #DominatedConvergenceTheorem #Lebesgue #MeasurableFunction #LebesgueFunction #LebesgueIntegration #RiemannIntegration #MeasureSpace

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes #Lebesgue integration more powerful than #Riemann integration. The theorem an be stated as follows:

Let \((f_n)\) be a sequence of measurable functions on a measure space \((\mathcal{S},\Sigma,\mu)\). Suppose that \((f_n)\) converges pointwise to a function \(f\) and is dominated by some Lebesgue integrable function \(g\), i.e. \(|f_n(x)|\leq g(x)\ \forall n\) and \(\forall x\in\mathcal{S}\). Then, \(f\) is Lebesgue integrable, and

\[\displaystyle\lim_{n\to\infty}\int_\mathcal{S}f_n\ \mathrm{d}\mu=\int_\mathcal{S}f\ \mathrm{d}\mu\]
#ConvergenceTheorem #Convergence #DominatedConvergenceTheorem #Lebesgue #MeasurableFunction #LebesgueFunction #LebesgueIntegration #RiemannIntegration #MeasureSpace

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes #Lebesgue integration more powerful than #Riemann integration. The theorem an be stated as follows:

Let \((f_n)\) be a sequence of measurable functions on a measure space \((\mathcal{S},\Sigma,\mu)\). Suppose that \((f_n)\) converges pointwise to a function \(f\) and is dominated by some Lebesgue integrable function \(g\), i.e. \(|f_n(x)|\leq g(x)\ \forall n\) and \(\forall x\in\mathcal{S}\). Then, \(f\) is Lebesgue integrable, and

\[\displaystyle\lim_{n\to\infty}\int_\mathcal{S}f_n\ \mathrm{d}\mu=\int_\mathcal{S}f\ \mathrm{d}\mu\]
#ConvergenceTheorem #Convergence #DominatedConvergenceTheorem #Lebesgue #MeasurableFunction #LebesgueFunction #LebesgueIntegration #RiemannIntegration #MeasureSpace

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes #Lebesgue integration more powerful than #Riemann integration. The theorem an be stated as follows:

Let \((f_n)\) be a sequence of measurable functions on a measure space \((\mathcal{S},\Sigma,\mu)\). Suppose that \((f_n)\) converges pointwise to a function \(f\) and is dominated by some Lebesgue integrable function \(g\), i.e. \(|f_n(x)|\leq g(x)\ \forall n\) and \(\forall x\in\mathcal{S}\). Then, \(f\) is Lebesgue integrable, and

\[\displaystyle\lim_{n\to\infty}\int_\mathcal{S}f_n\ \mathrm{d}\mu=\int_\mathcal{S}f\ \mathrm{d}\mu\]
#ConvergenceTheorem #Convergence #DominatedConvergenceTheorem #Lebesgue #MeasurableFunction #LebesgueFunction #LebesgueIntegration #RiemannIntegration #MeasureSpace

DOMINATED CONVERGENCE THEOREM
Lebesgue's dominated convergence theorem provides sufficient conditions under which pointwise convergence of a sequence of functions implies convergence of the integrals. It's one of the reasons that makes #Lebesgue integration more powerful than #Riemann integration. The theorem an be stated as follows:

Let \((f_n)\) be a sequence of measurable functions on a measure space \((\mathcal{S},\Sigma,\mu)\). Suppose that \((f_n)\) converges pointwise to a function \(f\) and is dominated by some Lebesgue integrable function \(g\), i.e. \(|f_n(x)|\leq g(x)\ \forall n\) and \(\forall x\in\mathcal{S}\). Then, \(f\) is Lebesgue integrable, and

\[\displaystyle\lim_{n\to\infty}\int_\mathcal{S}f_n\ \mathrm{d}\mu=\int_\mathcal{S}f\ \mathrm{d}\mu\]
#ConvergenceTheorem #Convergence #DominatedConvergenceTheorem #Lebesgue #MeasurableFunction #LebesgueFunction #LebesgueIntegration #RiemannIntegration #MeasureSpace

`It is also called the #Cantor ternary #function, the #Lebesgue function, Lebesgue's singular function..the Devil's staircase, the Cantor #staircase function, and the Cantor–Lebesgue function. Georg Cantor (1884) introduced the Cantor function and mentioned that Scheeffer pointed out that it was a counterexample to an extension of the fundamental #theorem of #calculus claimed by Harnack.`

https://en.wikipedia.org/wiki/Cantor_function

#devilsStaircase #fractal #cantorSet #topology #analysis