Is a measure for a sigma algebra determined by its values for a generator of the sigma algebra? If the value of a measure on any subset in a generator of a sigma algebra is known, will the measure for the sigma algebra also be uniquely determined? Thanks!
 A: Consider flipping two coins.  Let $A$ be the event that the first coin is heads, and $B$ the event that the second coin is heads.   $A$ and $B$ together generate the $\sigma$-algebra of all possible events.  Suppose we know that $P(A) = P(B) = 1/2$ (i.e. each coin is unbiased).  This is not enough information to determine whether the two coins are independent, so $P$ is not completely determined.
This is the same counterexample that I gave in this answer to another question.  In its notation, $P$ and $Q$ agree on the events in $\mathcal{L}$, and $\sigma(\mathcal{L}) = \mathcal{F}$, but $P \ne Q$.
A: Consider the $\sigma$-algebra given by all subsets of $\{a,b,c\}$.
It is generated by $A = \{a,b\}$ and $C = \{b,c\}$.
Let $\mu(A) = \mu(C) = 1$.
It could be that $\mu(\{a\}) = \mu(\{b\}) = \mu(\{c\}) = \frac{1}{2}$.
It could also be that $\mu(\{a\}) = \mu(\{c\}) = \frac{1}{3}$ and $\mu(\{b\}) = \frac{2}{3}$.
A: Let consider the above mentioned assertion due to  Davide Giraudo. 
Fact 1.  Let $μ$  a $\sigma$-finite measure on a measurable space $(X,S)$ , $\cal{A}$  an algebra which generates $S$  and $μ_1 ,μ_2$   two measures on $S$  such that for each $A\in\cal{A}$ , $μ_1 (A)=μ_2 (A)=μ(A)$. Then $μ_1 (B)=μ_2 (B)$  for each $B\in S$ .
Remark 1. Fact 1 , in general, is not true if $X$ is not covered by a countable family of elements of $\cal{A}$ which have finite $\mu$-measures.
Example 1. Let $N$ denotes a set of all natural munbers. Let $ {\bf R}^N $ be the topological vector  space of all real-valued
sequences equipped with the Tykhonoff topology. Let us denote by $
B({\bf R}^N) $ the $\sigma$-algebra of all Borel subsets in $ {\bf
R}^N $.
Let $ (a_i)_{i \in N} $ and $ (b_i)_{i \in N} $ be sequences of real numbers
such that
$$
( \forall i )( i \in N \rightarrow a_i < b_i ).
$$
We put
$$
 A_n={\bf R}_0 \times \cdots \times {\bf R}_n \times (\prod \limits_{i > n}\Delta_i)~,
$$
for $n \in N$, where
$$
(\forall i)( i \in N \rightarrow {\bf R}_i={\bf R}~ \& ~
\Delta_i=[a_i;b_i[).
$$
We put also
$$
\Delta=\prod_{i \in N}\Delta_i.
$$
For an arbitrary natural number $i \in N$, consider the Lebesgue measure
$ \mu_i $  defined on the space ${\bf R}_i$ and satisfying the
condition $\mu_i(\Delta_i)=1$. Let us denote by $\lambda_i$ the
normed Lebesgue measure defined on the interval $\Delta_i$.
For an  arbitrary $n \in N$, let us denote by $\nu_n$ the measure  defined by
$$
\nu_n= \prod \limits_{1 \le i \le n} \mu_i \times \prod\limits_{i
> n} \lambda_i,
$$
and by ${\overline{\nu}}_n$ the Borel measure in the space ${\bf
R}^N$  defined by
$$
 ( \forall X)(X \in B({\bf R}^N)  \rightarrow  {\overline{\nu}}_n(X)=
\nu_n(X \cap A_n)).
$$
Following [G.Pantsulaia , Invariant and quasiinvariant measures in infinite-dimensional topological vector spaces. Nova Science Publishers, Inc., New York, 2007. xii+234 pp.](see  Lemma 5.1, p. 93), for an arbitrary Borel set $X \subseteq {\bf
R}^N $ there exists a limit
$$
{\nu}_{\Delta}(X)= \lim \limits_{n \rightarrow \infty}
\overline{\nu}_n(X).
$$
Moreover, the functional ${\nu}_{\Delta}$ is a nontrivial
$\sigma$-finite measure  defined on the Borel $\sigma$-algebra
$B({\bf R}^N)$.
Let $\Delta^{(1)}=[0,1[^{N}$ and $\Delta^{(2)}=[2,3[^{N}$. Let consider $\cal{A}$-a class of subsets of ${\bf R}^N$ defined by
$$
{\cal{A}}=\{X \times R^{N \setminus \{1,\cdots,n\}}:X \in \cal{B}({\bf R}^n)~\&~n \in N\}.
$$
Obviously, $\cal{A}$ is an algebra of subsets of ${\bf R}^N$ which generates the Borel $\sigma$-algebra $B({\bf R}^N)$.
On the one hand, the measures $\mu_1:={\nu}_{\Delta^{(1)}}$  and $\mu_2:={\nu}_{\Delta^{(2)}}$ are agree on $\cal{A}$. In particular, $\mu_1(X \times R^{N \setminus \{1,\cdots,n\}})=\mu_2(X \times R^{N \setminus \{1,\cdots,n\}})=+\infty$ if $n$-dimensional Lebesgue measure of $X$ is positive and $=0$ if $n$-dimensional Lebesgue measure of $X$ is zero.
On the other hand, we have that $\mu_1(\Delta^{(2)})=0$ and $\mu_2(\Delta^{(2)})=1$.  
Example 2(Simple example) Let $X=[0,1[$ and $(x_k)_{k \in N}$ and $(y_k)_{k \in N}$ be two different everywere dense in $[0,1[$ sequences. Let $\cal{A}$ be a class of all subsets of $[0,1[$ every element of which is presented as a union of a finite family of left closed and right open subintervals  of $[0,1[$. Obviously, $\cal{A}$ is the algebra of subset of $[0,1[$ which generates the Borel $\sigma$-algebra $\cal{B}([0,1[)$. Let define two measures $\mu_1$ and $\mu_2$ as follows:
$\mu_1(X)=\# (\{x_k:k \in N\} \cap X)$ and $\mu_2(X)=\# (\{y_k:k \in N\} \cap X)$ for $X \in \cal{B}([0,1[)$, where $\#$ denotes the counting measure. Then $\mu_1$ and 
$\mu_2$ are agree on $\cal{A}$  but they are dfferent because $(x_k)_{k \in N}$ and $(y_k)_{k \in N}$ are different. 
The following assertion is valid.
Fact 2.  Let $μ$  a $\sigma$-finite measure on a measurable space $(X,S)$ , $\cal{A}$  an algebra which generates $S$ such that $X$ is covered by a countable family of elements of $\cal{A}$ which have finite $\mu$-measures. If $μ_1 ,μ_2$ are  two measures on $S$  such that for each $A\in\cal{A}$ , $μ_1 (A)=μ_2 (A)=μ(A)$ then $\mu_1(B)=\mu_2(B)$ for all $B \in S$. 
A: Consider the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$, and the class $\mathcal C:=\left\{(a,+\infty),a\in\mathbb R\right\}$. Then the class $\mathcal C$ generates $\mathcal B(\mathbb R)$. 
Consider the counting measure $\mu$ over $\mathcal B(\mathbb R)$, that is $\mu(B)=\begin{cases} 
\operatorname{card} A&\mbox{ if }A \mbox{ is finite, }\\\
+\infty&\mbox{ otherwise,}
\end{cases}$ 
and the Lebesgue measure. These measures have the same value over the elements of $\mathcal C$, but since for example $\{0\}\in\mathcal B(\mathbb R)$, and $\mu(\{0\})=1\neq \lambda(\{0\})=0$, these measure can't be the same.  
However, we can  show the following result: 

Let $\mu_1$ and $\mu_2$ be two $\sigma$-finite measures on a measurable space $(X,\mathcal S)$ and let $\mathcal A$ be an algebra which generates $\mathcal S$. Assume that for each $A\in\mathcal A$, $\mu_1(A)=\mu_2(A)$. 
Then $\mu_1(B)=\mu_2(B)$ for each $B\in\mathcal S$.

A: There is a famous example: A compact separable metric space, two different finite Borel measures on the space, but the two measures agree with each other on all of the balls for the metric.  
R. O. Davies, "Measures not approximable or not specifiable by means of balls."  Mathematika 18 (1971) 157--160
A: Caratheodory’s measure extension theorem.:
Theorem 1.41 (Caratheodory). Let $A ⊂ 2^Ω$ be a ring and let $µ$ be a $σ$-finite
premeasure on $A$. There exists a unique measure $µ'$ on $σ(A)$ such that $µ'(E) = µ(E)$
for all $E ∈ A$. Furthermore, $µ'$ is $σ$-finite.
p. 19 of http://www.math.unipd.it/~daipra/didattica/galileiana-15/Parte-I.pdf
(The same is given on p. 14 of here: http://www.mathematik.uni-kl.de/~wwwstoch-alt/skripte/basicmeasuretheory_skript.pdf and here: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem )
premeasure is a $σ$-additive set function $\mu$ to $[0,\infty]$ with $\mu(\emptyset)=0$.
Ring contains $\emptyset$ and is closed w.r.t. unions of two elements and complements.
