# Equality of measures on a generated $\sigma$-algebra

Let $\mathscr{S}$ be a nonempty Set. Let $\mathscr{E}\subset \mathscr{P}( \mathscr{S})$ be the generator of the $\sigma$-algebra $\sigma (\mathscr{E}) =:\mathscr{A}$. Let $\mu$ and $\mu’$ be measures on $\mathscr{A}$ and $\mu(\mathscr{S})=\mu'(\mathscr{S})$ and if necessary $\mu(\mathscr{S})<\infty$

Is the following statement true:
$\Big(\forall A\in\mathscr{E}:\mu(A)=\mu'(A)\Big)\Rightarrow \Big(\forall A \in \mathscr{A}:\mu(A)=\mu'(A)\Big)$

#### Solutions Collecting From Web of "Equality of measures on a generated $\sigma$-algebra"
However there are some interesting results. All the them can be proved using what seem to be the two standard tools to prove these kinds of results: Dynkin’s $\pi-\lambda$ Theorem and the Monotone Class Theorem. Both are not very different, I might have read that the first one is more used in probability theory and the latest in measure theory for some reason.
$1)$ If two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system) and satisfy that $\mu_1(X)=\mu_2(X)<\infty$, then $\mu_1=\mu_2$ on the $\sigma$-algebra generated by $\mathcal{A}$ (see a proof sketch here, under the Theorem; basically it is enough to show that $\{A\in \sigma(\mathcal{A}): \mu_1(A)=\mu_2(A)\}$ is a $\lambda$-system). For example, if two finite Borel measures on any topological space coincide on all open sets, then they coincide on all Borel sets.
$2)$ [Generalization of $1$] Suppose two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system). Moreover suppose there exists a countable nested subfamily of sets $\mathcal{B}\subseteq \mathcal{A}$ that cover $X$ and have finite $\mu_1$-(and $\mu_2$-) measure. Then $\mu_1=\mu_2$ in the $\sigma$-algebra generated by $\mathcal{A}$ (proof scheme similar to $1$, essentially a straightforward generalization of the answer here). So for example the Lebesgue measure in all Borel sets in $\mathbb{R}^n$ is uniquely determined by its values on boxes/intervals.
$3)$ Up to a multiplicative constant, Lebesgue measure is the only translation-invariant measure on the Borel sets that puts finite measure on the unit interval. This can be generalised to higher dimensions considering the unit box $[0,1]^n$ instead of the unit interval. You can find proofs here using the $\pi – \lambda$ Theorem. A proof using the Monotone Class Theorem also seems quite straightforward.