A measure is completely determined by its values on borel sets? Let us assume that we have two probability measures defined on a compact set $X\subset \mathbb{R}^d$ with a $\sigma$-algebra which properly contains Borel sets of X (considering the topology given by the usual metric). My question is: if both measures coincide on Borel sets, can be sure that they also coincide on the whole bigger $\sigma$-algebra?
I think that in some specific examples this is clear (for instance, considering Lebesgue measurable sets versus Borel sets), but I am not sure about what happens in contexts where maybe the bigger $\sigma$-algebra is not the completion of the Borel $\sigma$-algebra.
Being more specific, my question comes while reading the book Discrete Energy on rectifiable sets, where the authors define Borel measures as measures containing Borel sets in its $\sigma$-algebra (but accepting for instance the Lebesgue measure as a Borel measure). In page 135 they consider a positive lower-semicontinuous function $K(x)$ and write the following inequality (I am simplifying a bit the situation):
$$\int K(x)d\mu_1(x) \leq \int K \mu_2(x),$$
where $\mu_1,\mu_2$ represents two "Borel measures" that have checked to be $\mu_1\leq \mu_2$ just for Borel sets, not all the measurable sets of our $\sigma$-algebra.
I don't think the usual arguments like motonone class theorem or Dynkin systems apply here, since the Borel sets wouldn't generate the whole $\sigma$-algebra, and at the same time I am not sure about if we also can think that the the measure on the bigger $\sigma$-algebra is some kind of completion of the smaller one.
Thanks in advance.
 A: The question is answered in the negative by
the following argument, which is based closely on Donald L. Cohn,
Measure Theory (2nd ed. 2013), Exercises 1.5.11-12.
(I may have made heavy weather of it.  Suggestions for
simplifications would be gratefully received. Cohn is
not to blame for my choices of notation.) Another
good reference may be R. M. Dudley, Real Analysis and
Probability (2nd ed. 2002), Theorems 3.3.6 and 3.4.4 (the
latter is credited to van Vleck, 1908), and Problems 3.4.3-5, but I
haven't worked though these. It would probably also be worth working
through the development in H. L. Royden, Real Analysis (3rd
ed. 1988), because Theorem 38 of Chapter 12 suggests that there is
never a need to assume that $\mu$ is finite.
Let $(X, \mathscr{A}, \mu)$ be a measure space.  As usual, for any
set $B \subseteq X$ we write
\begin{gather*}
\mu^*(B) = \inf\{\mu(A) : B \subseteq A \in \mathscr{A}\}, \\
\mu_*(B) = \sup\{\mu(A) : B \supseteq A \in \mathscr{A}\}.
\end{gather*}
Recall that for all $B \subseteq X,$ there exist $\mu$-measurable
sets $B^*$ and $B_*$ such that $B_* \subseteq B \subseteq B^*,$
$\mu(B^*) = \mu^*(B),$ and $\mu(B_*) = \mu_*(B).$ These sets are
generally not unique (and this notation for them is not standard).
Also, for all $B \subseteq X,$
\begin{equation}
\label{4383954:eq:1}\tag{1}
\mu^*(B) + \mu_*(X - B) = \mu(X).
\end{equation}
Suppose $\mu^*(B) < +\infty.$ If $A \in \mathscr{A}$ and
$A \cap B = \emptyset,$ then $B \subseteq B^* - A,$ therefore
$$
\mu(B^*) = \mu^*(B) \leqslant \mu(B^* - A) =
\mu(B^*) - \mu(A \cap B^*),
$$
therefore $\mu(A \cap B^*) = 0.$ Suppose now that
$A_1, A_2 \in \mathscr{A}$ and $A_1 \cap B = A_2 \cap B.$ Then
$(A_1 \bigtriangleup A_2) \cap B =
(A_1 \cap B) \bigtriangleup (A_2 \cap B) = \emptyset,$ therefore
$$
\mu((A_1 \cap B^*) \bigtriangleup (A_2 \cap B^*)) =
\mu((A_1 \bigtriangleup A_2) \cap B^*) = 0,
$$
therefore $\mu(A_1 \cap B^*) = \mu (A_2 \cap B^*).$ We can therefore
define a function
$$
\mu_B \colon \mathscr{A}_B \to [0, \infty), \
A \cap B \mapsto \mu(A \cap B^*),
$$
where
$$
\mathscr{A}_B = \{A \cap B : A \in \mathscr{A}\}.
$$
It is clear that $\mathscr{A}_B$ is a $\sigma$-algebra on $B.$
(It is called the trace of $\mathscr{A}$ on $B.$) Obviously
$\mu_B(\emptyset) = 0.$ To prove that $\mu_B$ is countably additive,
and therefore a measure on $\mathscr{A}_B,$ we need a lemma:
Lemma 1. If $(X, \mathscr{A}, \mu)$ is a measure space, and
$(A_n)_{n \geqslant 1}$ is a sequence in $\mathscr{A}$ such that
$\mu(A_m \cap A_n) = 0$ whenever $m \ne n,$ then
$$
\mu\left(\bigcup_{n=1}^\infty A_n\right) =
\sum_{n=1}^\infty \mu(A_n).
$$
Proof. For all $n,$ define $C_n = \bigcup_{m < n}A_m.$ Then
$$
\mu(A_n \cap C_n) = \mu\bigg(\bigcup_{m < n}(A_n \cap A_m)\bigg)
\leqslant \sum_{m < n} \mu(A_n \cap A_m) = 0,
$$
therefore $\mu(A_n \cap C_n) = 0,$ therefore
$$
\mu(A_n) = \mu(A_n \cap C_n) + \mu(A_n - C_n) = \mu(A_n - C_n).
$$
The sets $A_n - C_n$ are pairwise disjoint, and their
union is $\bigcup_{n=1}^\infty A_n,$ therefore
$$
\mu\left(\bigcup_{n=1}^\infty A_n\right) =
\mu\left(\bigcup_{n=1}^\infty (A_n -  C_n)\right) =
\sum_{n=1}^\infty \mu(A_n - C_n) =
\sum_{n=1}^\infty \mu(A_n). \quad \square
$$
Now, let $(D_n)_{n \geqslant 1}$ be an infinite sequence of disjoint
sets that belong to $\mathscr{A}_B.$ By the axiom of countable
choice, there exists a sequence $(A_n)_{n \geqslant 1}$ in
$\mathscr{A}$ such that $D_n = A_n \cap B$ for all $n.$ For all
$m, n$ such that $m \ne n,$
$$
(A_m \cap A_n) \cap B = D_m \cap D_n = \emptyset,
$$
therefore (by the first result proved in the paragraph preceding the
lemma)
$$
\mu((A_m \cap B^*) \cap (A_n \cap B^*)) =
\mu((A_m \cap A_n) \cap B^*) = 0.
$$
Applying the lemma to the sequence of sets
$(A_n \cap B^*)_{n \geqslant 1},$ we get
\begin{multline*}
\mu_B\left(\bigcup_{n=1}^\infty D_n\right) =
\mu_B\left(\left(\bigcup_{n=1}^\infty A_n\right) \cap B\right) =
\mu\left(\left(\bigcup_{n=1}^\infty A_n\right) \cap B^*\right) \\
= \mu\left(\bigcup_{n=1}^\infty(A_n \cap B^*)\right) =
\sum_{n=1}^\infty \mu(A_n \cap B^*) = \sum_{n=1}^\infty \mu_B(D_n).
\end{multline*}
This completes the proof that $(B, \mathscr{A}_B, \mu_B)$ is a
measure space, for all $B \subseteq X$ such that
$\mu^*(B) < +\infty.$
The definition of $\mu_B$ involves the choice of a set $B^*$ such
that $B \subseteq B^* \in \mathscr{A}$ and $\mu(B^*) = \mu^*(B).$
The next result shows that it does not matter which such set is
chosen:
Proposition 2. $\mu_B(C) = \mu^*(C),$ for all
$C \in \mathscr{A}_B.$
Proof. There exists $A \in \mathscr{A}$ such that
$C = A \cap B.$ Let $C^*$ be such that
$C \subseteq C^* \in \mathscr{A}$ and $\mu(C^*) = \mu^*(C).$ Then
$C = C \cap C^*,$ therefore
$$
C = (A \cap B) \cap C^* = (A \cap C^*) \cap B.
$$
Therefore, by the definition of $\mu_B,$
$$
\mu_B(C) = \mu((A \cap C^*) \cap B^*) =
\mu((A \cap B^*) \cap C^*) \leqslant \mu(C^*) = \mu^*(C).
$$
But also $\mu_B(C) = \mu(A \cap B^*),$ and $C \subseteq A \cap B^*,$
therefore $\mu_B(C) \geqslant \mu^*(C).$ So $\mu_B(C) = \mu^*(C). \
\square$
If $X$ is a set, and $\mathscr{C}$ is any collection of subsets of
$X,$ then there is a smallest $\sigma$-algebra on $X$ that contains
$\mathscr{C}.$ We denote it by $\sigma(\mathscr{C}).$ It is easy
to show that if $\mathscr{A}$ is a $\sigma$-algebra on $X,$ and $E$
is a subset of $X,$ and $F = X - E,$ then
$$
\sigma(\mathscr{A} \cup \{E\}) = \sigma(\mathscr{A} \cup \{F\}) =
\{(A_1 \cap E) \cup (A_2 \cap F) : A_1, A_2 \in \mathscr{A}\}.
$$
From now on, let $(X, \mathscr{A}, \mu)$ be a finite measure
space. [As noted above, the assumption of finiteness may in fact be
unnecessary; but I don't know how to prove this; and in any case we
needn't worry about it for the present question.] Let $E$ and
$F$ be as above, and let $E^*$ and $E_*$ be $\mu$-measurable sets
satisfying $E_* \subseteq E \subseteq E^*,$ $\mu(E^*) = \mu^*(E)$
and $\mu(E_*) = \mu_*(E).$ Let $F^* = X - E_*,$ and $F_* = X - E_*.$
Then $F_* \subseteq F \subseteq F^*,$ and by \eqref{4383954:eq:1},
$\mu(F^*) = \mu^*(F)$ and $\mu(F_*) = \mu_*(F).$
Lemma 3. If $(X, \mathscr{A}, \mu)$ is a measure space,
then for all $Y \in \mathscr{A}$ the function
$$
\hat{\mu}_Y \colon \mathscr{A} \to [0, +\infty],
\ A \mapsto \mu(A \cap Y)
$$
is a measure on $\mathscr{A},$ and
$$
\mu = \hat{\mu}_Y + \hat{\mu}_{X - Y}. \quad \square
$$
In particular,
\begin{align*}
\mu & = \hat{\mu}_{E_*}\! + \hat{\mu}_{F^*}
\\  & = \hat{\mu}_{F_*} + \hat{\mu}_{E^*}\!.
\end{align*}
We shall use these two decompositions to extend the function
$\mu \colon \mathscr{A} \to [0, +\infty)$ in two ways to the larger
$\sigma$-algebra $\mathscr{E} = \sigma(\mathscr{A} \cup \{E\}).$
For all $B \in \mathscr{E},$ define
\begin{align*}
\mu_0(B) & = \mu(B \cap E_*) + \mu_F(B \cap F), \\
\mu_1(B) & = \mu(B \cap F_*) + \mu_E(B \cap E).
\end{align*}
Because $E_*, F_* \in \mathscr{E},$ the first term on the right hand
side of each of these two equations is the value of a measure,
defined by Lemma 3 in the same way as for $\mu,$ but on the larger
$\sigma$-algebra $\mathscr{E}.$ The clever part is the second term,
because as observed above, we have
$B = (A_1 \cap E) \cup (A_2 \cap F)$ for some
$A_1, A_2 \in \mathscr{A},$ whence
\begin{align*}
\mu_0(B) & = \mu(A_1 \cap E_*) + \mu(A_2 \cap F^*), \\
\mu_1(B) & = \mu(A_2 \cap F_*) + \mu(A_1 \cap E^*).
\end{align*}
If $B = E,$ we can take $A_1 = X$ and $A_2 = \emptyset$; and if
$B = F,$ we can take $A_2 = X$ and $A_1 = \emptyset.$ Hence
\begin{gather*}
\mu_0(E) = \mu(E_*) = \mu_*(E), \quad
\mu_0(F) = \mu(F^*) = \mu^*(F), \\
\mu_1(E) = \mu(E^*) = \mu^*(E), \quad
\mu_1(F) = \mu(F_*) = \mu_*(F).
\end{gather*}
Therefore, if $\mu_*(E) < \mu^*(E),$ i.e., if $E$ is not
$\mu$-measurable, then $\mu_0 \ne \mu_1,$ and the measure $\mu$ on
$\mathscr{A}$ has been extended to two distinct measures on the
larger $\sigma$-algebra $\mathscr{E}.$ In fact, if $\alpha$ is any
number in the interval $[\mu_*(E), \mu^*(E)],$ then by defining
$$
t = \frac{\mu^*(E) - \alpha}{\mu^*(E) - \mu_*(E)}, \quad
1 - t = \frac{\alpha - \mu_*(E)}{\mu^*(E) - \mu_*(E)}
$$
and
$$
\nu(B) = t\mu_0(B) + (1 - t)\mu_1(B) \quad (B \in \mathscr{E}),
$$
we have succeeded in extending $\mu$ to a measure $\nu$ on
$\mathscr{E}$ such that $\nu(E) = \alpha.$
We quote without proof the result that, if $\lambda$ denotes the
restriction of Lebesgue measure on $\mathbb{R}$ to the closed
interval $X = [0, 1],$ then there exists a set $E \subset X$ such
that $\lambda_*(E) = 0$ and $\lambda^*(E) = 1.$ (The proof is
neither long nor difficult, but all its details are irrelevant here.
The result follows easily from Proposition 1.4.11 in Cohn, and is
stated explicitly as Theorem 3.4.4 in Dudley.)
It follows that for every number $\alpha$ between $0$ and $1$
inclusive, there exists a probability space $(X, \mathscr{E}, \nu),$
where $X = [0, 1] \subset \mathbb{R},$ $\mathscr{E}$ (the same
$\sigma$-algebra for every value of $\alpha$) contains the Borel
sets of $X$ (which are the intersections with $X$ of the Borel sets
of $\mathbb{R}$), $\nu(B) = \lambda(B)$ for every Borel set $B,$ and
$\nu(E) = \alpha.$
