will make M symmetric? Show all working.
be a measure space, and 
.![$$f:E\to[0,\infty]$$](http://latex.artofproblemsolving.com/b/9/2/b92b380e3810cd7880fcc9f6b0940fcb8994c8f4.png)
be a measurable function, and 
.
such that 
.
Y by
For which integers 
does an infinite string of characters
![\[ S = c_{1} \cdot c_{2} \cdot c_{3} \cdot c_{4} \cdot c_{5} \ldots \]](//latex.artofproblemsolving.com/e/e/7/ee7ef9fbfae034fda276a995bcb82757e945171a.png)
exist such that for all
there are exactly 
distinct substrings of 
with length 
? (A substring is a finite subsequence of consecutive characters.)
Does an infinite string exist such that there are exactly 
distinct substrings of with length ? is the th prime.
does an infinite string of characters![\[ S = c_{1} \cdot c_{2} \cdot c_{3} \cdot c_{4} \cdot c_{5} \ldots \]](http://latex.artofproblemsolving.com/e/e/7/ee7ef9fbfae034fda276a995bcb82757e945171a.png)
exist such that for all
there are exactly 
distinct substrings of 
with length 
? (A substring is a finite subsequence of consecutive characters.)
Does an infinite string exist such that there are exactly 
distinct substrings of with length ? is the th prime.
and it is called the complexity function of the word. Just a few remarks.
for all 
and the word is constant. Hence not possible. More generally, if 
for some 
then the word is ultimately periodic and 
nice examples can be found in a paper by Günter Rote, Journal of Number Theory 46. I will quote Theorem 2 here.
with 
and 
for all 
as follows: 
.
for all 
be a positive integer. In my thesis I construct words 
such that 
for 
.
for all 
(with 
)
and 
.
.
