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Suppose $(E, \mathcal{A})$ is a measurable space. Let $\mu$ and $\gamma$ be two distinct measures of this space. Now we say that $\gamma$ is absolutely continuous with respect to $\mu$ if for every $A \in \mathcal{A} , \mu(A) = 0 \Rightarrow \gamma(A) = 0 $.

I am unable to understand the essence of this definition of "absolutely continuous". I mean what is so absolute and continuous about these measures?

Dark_Knight
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    Are you asking to understand what the name has to do with the concept, or are you asking how to understand the concept itself? If the latter, can you say more about exactly what you don't understand about the concept? (You don't need to understand the name to understand the concept; the name is largely a historical relic that is not especially useful for understanding the concept.) – Eric Wofsey Mar 13 '16 at 05:27
  • @ Eric Wofsey - Well, I was doubtful about the name of the concept. The definition seems to be fine! – Dark_Knight Mar 13 '16 at 05:32
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    Also its good if one understands why any concept is named so as it many times gives motivation for the concepts! – Dark_Knight Mar 13 '16 at 06:18

2 Answers2

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For simplicity, assume that $\gamma$ is a finite measure. One can show that this condition is equivalent to the following: For each $\varepsilon > 0$ there exists some $\delta > 0$ such that $\mu(A)<\delta$ implies $\gamma(A) < \varepsilon$. Hence, $\gamma$ can be regarded continuous with respect to $\mu$. In this case, there exists a measurable function $f$ such that $\gamma(A) = \int_A f\,d\mu$. If the measure space is $\mathbb R$ and $\mu$ is the Lebesgue measure, then $\gamma((-\infty,x]) = \int_{-\infty}^xf(t)\,dt$ and this function (in $x$) is absolutely continuous.

Friedrich Philipp
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Absolute continuity of measures is named after absolutely continuous functions. A function $f:[a,b]\to\mathbb{R}$ is called absolutely continuous if for each $\epsilon>0$ there exists a $\delta>0$ such that for any finite collection of disjoint intervals $(x_1,y_1),(x_2,y_2),\dots,(x_n,y_n)\subset[a,b]$ whose total length is less than $\delta$, $$\sum_{k=1}^n |f(y_k)-f(x_k)|<\epsilon.$$

Note that if you restrict to the case $n=1$, this is just the usual definition of uniform continuity, which is equivalent to continuity since $[a,b]$ is compact. So this is somehow a stronger version of continuity, which might be reasonable to call "absolute continuity" (though I'm not sure exactly what's "absolute" about it).

So, what does this have to do with measures? Well, it turns out that a measure $\gamma$ on $[a,b]$ is absolutely continuous with respect to Lebesgue measure iff the function $f(c)=\gamma([a,c])$ is absolutely continuous. So in general, the relationship between absolutely continuous measures is a generalization of the relationship between absolutely continuous functions (for which the name makes a bit more sense) and Lebesgue measure.

Eric Wofsey
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