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Absolutely Continuous


A measure lambda is absolutely continuous with respect to a measure mu if lambda(E)=0 for every measurable set E with mu(E)=0. This makes sense as long as mu is a positive measure, such as Lebesgue measure, but lambda can be any measure, possibly a complex measure.

If mu is a sigma-finite positive measure and lambda is a finite complex measure, the Radon-Nikodym theorem makes this equivalent to saying that

 lambda(E)=int_Efdmu,

where the integral is the Lebesgue integral, for some integrable function f. The function f is like a derivative, and is called the Radon-Nikodym derivative dlambda/dmu. Here, sigma-finiteness means that the underlying space is a countable union of measurable sets of finite mu-measure. The finiteness assumption on lambda includes finite positive measures and signed measures.

The measure supported at 0 (mu(E)=1 iff 0 in E) is not absolutely continuous with respect to Lebesgue measure, and is a singular measure.


See also

Absolutely Continuous Function, Complex Measure, Concentrated, Haar Measure, Lebesgue Decomposition, Lebesgue Measure, Mutually Singular, Singular Measure

This entry contributed by Todd Rowland

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References

Hunter, J. K. "Radon-Nikodym Theorem" and "Complex Measures." §6.8 and 6.9 in Measure Theory. 2011. https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes.pdf.

Referenced on Wolfram|Alpha

Absolutely Continuous

Cite this as:

Weisstein, Eric W., with contributions by Todd Rowland. "Absolutely Continuous." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/AbsolutelyContinuous.html

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