Let $\mu$ and $\lambda$ be two measures on a $\sigma$-algbra $\mathfrak{F}$ on $\Omega$, such that $\mu (A)=\lambda(A)$ for any $A\in \mathfrak C$, where $\mathfrak C\subset\mathfrak{F}$ is a collection having property if $A$ and $B$ are in $\mathfrak C$, then so is $A\cap B$. Assume there is $A_i\in \mathfrak C$, such that $\bigcup_{i=1}^\infty A_i=\Omega$ and $\mu(A_i)<\infty$ for all $i$. Then prove $\{A\in\sigma(\mathfrak C), \mu(A)=\lambda(A)\}$ is a $\sigma$-algebra.

This is one question from my homework. At first sight, it's easy. But actually, I have tried much time. I doubt that this question might be wrong.

  • $\begingroup$ I believe, it suffices to show that the family is a Dynkin system. Usually, this statement is used to prove the "uniqueness-of-measures-theorem" - and in this case, Dynkin system is enough. $\endgroup$
    – saz
    Aug 20, 2014 at 14:37
  • $\begingroup$ Yes. The key point is to prove $\{A\in\sigma(\mathfrak C), \mu(A)=\lambda(A)\}$ is $\lambda-$system. However, It's not easy. $\endgroup$
    – Shine
    Aug 20, 2014 at 14:55

1 Answer 1



$$\mathcal{D} := \{A \in \mathcal{F}; \mu(A)=\lambda(A)\}.$$

  1. $\emptyset \in \mathcal{D}$: This follows obviously from the fact that $\mu(\emptyset) = \lambda(\emptyset)=0$.
  2. Let $A \in \mathcal{D}$. By assumption, we can choose a sequence of sets $(G_n)_n \subseteq \mathcal{F}$ such that $G_n \uparrow \Omega$, $\mu(G_n)=\lambda(G_n)<\infty$. (Hint: $G_n := \bigcup_{j=1}^n A_j$.) Then $$\begin{align*} \mu(G \cap A^c) &= \mu(G_n \backslash A) = \mu(G_n)-\mu(G_n \cap A) \\ &= \lambda(G_n) - \lambda(G_n \cap A) = \lambda(G_n \backslash A) \\ &= \lambda(G_n \cap A^c). \end{align*}$$ It follows from the continuity of the measures that $\mu(A^c) = \lambda(A^c)$. Hence, $A^c \in \mathcal{D}$.
  3. Let $(C_j)_j \subseteq \mathcal{D}$ be a sequence of disjoint sets. Using the $\sigma$-additivity of $\mu$ and $\lambda$, it is not difficult to see that $$\mu\left( \bigcup_j C_j \right) = \lambda \left( \bigcup_j C_j \right),$$ i.e. $\bigcup_j C_j \in \mathcal{D}$.

This shows that $\mathcal{D}$ is a Dynkin-system. As $\mathcal{C}$ is $\cap$-stable, this implies in particular $\sigma(\mathcal{C}) \subseteq \mathcal{D}$, i.e. $\lambda(A) = \mu(A)$ for all $A \in \sigma(\mathcal{C})$.

  • $\begingroup$ Almost done. However, I have one qustion. we can't choose $G_n \uparrow\Omega$ by assumption. We should cheak this carefully. $\endgroup$
    – Shine
    Aug 20, 2014 at 15:15
  • $\begingroup$ @Shine Why not? Set $G_n := \bigcup_{j=1}^n A_j$. $\endgroup$
    – saz
    Aug 20, 2014 at 15:20
  • $\begingroup$ This $G_n$ may not be in $\mathfrak C$ since it's just closed under intersection. $\endgroup$
    – Shine
    Aug 20, 2014 at 15:47
  • $\begingroup$ @Shine They are in $\mathcal{C}$ - I didn't claim that this it is that obvious. Do you know a formula for the measure of a finite union of sets? Hint: $$\mu(A \cup B) = \mu(A)+\mu(B) - \mu(A \cap B)$$ (if $\mu(A \cap B)<\infty$). $\endgroup$
    – saz
    Aug 20, 2014 at 17:35
  • $\begingroup$ Yes. Thank you very much. So you figured it out. Remind: $G_n\in \mathfrak D$, not $G_n\in \mathfrak C$. $\endgroup$
    – Shine
    Aug 21, 2014 at 1:55

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