Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Absolute continuity
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Relation between the two notions of absolute continuity== A finite measure ''μ'' on [[Borel set|Borel subsets]] of the real line is absolutely continuous with respect to [[Lebesgue measure]] if and only if the point function: :<math>F(x)=\mu((-\infty,x])</math> is an absolutely continuous real function. More generally, a function is locally (meaning on every bounded interval) absolutely continuous if and only if its [[distributional derivative]] is a measure that is absolutely continuous with respect to the Lebesgue measure. If absolute continuity holds then the Radon–Nikodym derivative of ''μ'' is equal almost everywhere to the derivative of ''F''.<ref>{{harvnb|Royden|1988|loc=Problem 12.17(b) on page 303}}.</ref> More generally, the measure ''μ'' is assumed to be locally finite (rather than finite) and ''F''(''x'') is defined as ''μ''((0,''x'']) for {{nowrap|''x'' > 0}}, 0 for {{nowrap|1=''x'' = 0}}, and −''μ''((''x'',0]) for {{nowrap|''x'' < 0}}. In this case ''μ'' is the [[Lebesgue–Stieltjes integration|Lebesgue–Stieltjes measure]] generated by ''F''.<ref>{{harvnb|Athreya|Lahiri|2006|loc=Sect. 1.3.2, page 26}}.</ref> The relation between the two notions of absolute continuity still holds.<ref>{{harvnb|Nielsen|1997|loc=Proposition 15.7 on page 252}}; {{harvnb|Athreya|Lahiri|2006|loc=Theorem 4.4.3 on page 131}}; {{harvnb|Royden|1988|loc=Problem 12.17(a) on page 303}}.</ref>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)