Return to Answer

forgot logs

Source Link

edited May 20, 2011 at 2:42

6k
2
52
57

There's a very good notion of "local" dimension of a measure at a point $x$: $$ \dim_x(\mu) = \lim_{r\rightarrow 0}\frac{\mu(B_r(x))}{r}$$$$ \dim_x(\mu) = \lim_{r\rightarrow 0}\frac{\log\mu(B_r(x))}{\log r}$$

where $B_r(x)$ is the ball of radius $r$ centered at $x$. (Intuitively, we expect that in a $d$-dimensional space, the volume of a ball is proportional to the $d$^th power of the radius, which immediately leads to this definition.) In general, the local dimension isn't defined everywhere and depends on $x$ when it is, but under certain conditions, it is constant $\mu$-almost everywhere, in which case it makes sense to call it the dimension of the measure.

In many cases, the measure $\mu$ is more interesting than its support, and the dimension defined thusly will reflect this. For example, consider a stochastic map on the interval $[0,1]$ that maps it affinely onto $[0,1/4]$, $[1/4,3/4]$, or $[3/4,1]$, each with probability 1/3. If $\mu$ is the invariant measure, then the support of $\mu$ is the whole interval, but you can check that $\dim_x(\mu)=\frac{3\log 3}{5\log 2}$ for $\mu$-a. e. $x$ (if I didn't screw it up). You can also verify that there are plenty of exceptional $x$ for which $\dim_x(\mu)$ is something else or undefined.

There's a very good notion of "local" dimension of a measure at a point $x$: $$ \dim_x(\mu) = \lim_{r\rightarrow 0}\frac{\mu(B_r(x))}{r}$$

where $B_r(x)$ is the ball of radius $r$ centered at $x$. (Intuitively, we expect that in a $d$-dimensional space, the volume of a ball is proportional to the $d$^th power of the radius, which immediately leads to this definition.) In general, the local dimension isn't defined everywhere and depends on $x$ when it is, but under certain conditions, it is constant $\mu$-almost everywhere, in which case it makes sense to call it the dimension of the measure.

In many cases, the measure $\mu$ is more interesting than its support, and the dimension defined thusly will reflect this. For example, consider a stochastic map on the interval $[0,1]$ that maps it affinely onto $[0,1/4]$, $[1/4,3/4]$, or $[3/4,1]$, each with probability 1/3. If $\mu$ is the invariant measure, then the support of $\mu$ is the whole interval, but you can check that $\dim_x(\mu)=\frac{3\log 3}{5\log 2}$ for $\mu$-a. e. $x$ (if I didn't screw it up). You can also verify that there are plenty of exceptional $x$ for which $\dim_x(\mu)$ is something else or undefined.

There's a very good notion of "local" dimension of a measure at a point $x$: $$ \dim_x(\mu) = \lim_{r\rightarrow 0}\frac{\log\mu(B_r(x))}{\log r}$$

where $B_r(x)$ is the ball of radius $r$ centered at $x$. (Intuitively, we expect that in a $d$-dimensional space, the volume of a ball is proportional to the $d$^th power of the radius, which immediately leads to this definition.) In general, the local dimension isn't defined everywhere and depends on $x$ when it is, but under certain conditions, it is constant $\mu$-almost everywhere, in which case it makes sense to call it the dimension of the measure.

In many cases, the measure $\mu$ is more interesting than its support, and the dimension defined thusly will reflect this. For example, consider a stochastic map on the interval $[0,1]$ that maps it affinely onto $[0,1/4]$, $[1/4,3/4]$, or $[3/4,1]$, each with probability 1/3. If $\mu$ is the invariant measure, then the support of $\mu$ is the whole interval, but you can check that $\dim_x(\mu)=\frac{3\log 3}{5\log 2}$ for $\mu$-a. e. $x$ (if I didn't screw it up). You can also verify that there are plenty of exceptional $x$ for which $\dim_x(\mu)$ is something else or undefined.

Source Link

created May 20, 2011 at 2:31

6k
2
52
57

There's a very good notion of "local" dimension of a measure at a point $x$: $$ \dim_x(\mu) = \lim_{r\rightarrow 0}\frac{\mu(B_r(x))}{r}$$

where $B_r(x)$ is the ball of radius $r$ centered at $x$. (Intuitively, we expect that in a $d$-dimensional space, the volume of a ball is proportional to the $d$^th power of the radius, which immediately leads to this definition.) In general, the local dimension isn't defined everywhere and depends on $x$ when it is, but under certain conditions, it is constant $\mu$-almost everywhere, in which case it makes sense to call it the dimension of the measure.

In many cases, the measure $\mu$ is more interesting than its support, and the dimension defined thusly will reflect this. For example, consider a stochastic map on the interval $[0,1]$ that maps it affinely onto $[0,1/4]$, $[1/4,3/4]$, or $[3/4,1]$, each with probability 1/3. If $\mu$ is the invariant measure, then the support of $\mu$ is the whole interval, but you can check that $\dim_x(\mu)=\frac{3\log 3}{5\log 2}$ for $\mu$-a. e. $x$ (if I didn't screw it up). You can also verify that there are plenty of exceptional $x$ for which $\dim_x(\mu)$ is something else or undefined.