The dimension of the measure mu is the smallest dimension of among all
sets A with mu(A) > 0. |
We shall show that the value of f(alpha) satisfying f(alpha) = alpha
is the dimension of the measure mu. This will take several steps. |
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(1) The diagonal line intersects the graph of f(alpha) exactly at one point,
and that point corresponds to q = 1. The proof involves
showing tau(1) = 0. |
Then for q = 1 from alpha = - dtau/dq we have
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f(alpha) = alpha = (p1ln(p1) + ... + pNln(pN))/(p1ln(r1)
+ ... + pNln(rN)) |
We show that this expression is the smallest dimension of any set A with
mu(A) > 0. |
First, note the region S(i1...ik) with address
i1...ik has measure |
mu(S(i1...ik)) = p1n1(k) ...
pNnN(k) |
where n1(k) is the number of 1s in i1...ik, n2(k) is the number of 2s
in i1...ik, and so on. |
For almost all sequences i1i2..., nj(k)/k -> pj
as k -> infinity. Consequently, |
(1/k)ln(mu(S(i1...ik))/diam(S(i1...ik))s) |
= (1/k)ln((p1n1(k) ...
pNnN(k))/(ri1...rik)s) |
= (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) +
(s/k)ln(1/(ri1...rik)) |
<= (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) +
(s/k)ln(1/(min{ri})k) |
= (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) +
s*ln(1/(min{ri})) |
-> p1ln(p1) + ... + pNln(pN) +
s*ln(1/(min{ri})) |
|
as k -> infinity. Consequently, if s < -(p1ln(p1) + ... +
pNln(pN))/ln(1/min{ri}) then |
limk -> infinity(1/k)ln(mu(S(i1...ik))/diam(S(i1...ik))s) < 0 |
so limk -> infinityln(mu(S(i1...ik))/diam(S(i1...ik))s) = -infinity |
and limk -> infinitymu(S(i1...ik))/diam(S(i1...ik))s = 0 |
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From this we see Hs(A) >= mu(A)/c for all c > 0. |
Because mu(A) > 0, we see Hs(A) = infinity and
so d(A) >= s. |
Then for all sets A with mu(A) > 0, |
d(A) | >= -(p1ln(p1) + ... +
pNln(pN))/ln(1/min{ri}) |
| >= (p1ln(p1) + ... + pNln(pN))/(p1ln(r1)
+ ... + pNln(rN)) |
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