Show that if $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing sequence of measurable sets (so $A_j\subseteq A_{j+1}$ for every positive integer $j$), then we have $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}m(A_j)$$

Here is my proof:

According to the $\sigma$-algebra property, $\bigcup_{j=1}^{\infty}A_j$ is a measurable set, so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$.

Firstly I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. This is because for any given positive integer $N$, $A_N\subseteq \bigcup_{j=1}^{\infty}A_i$, according to monotonicity, we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$.

Secondly I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$. For any given positive integer $N$, $\bigcup_{j=1}^N A_j = A_N$. According to monotonicity,we have $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$.

Combine the above two arguments ,we will see that $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty} m(A_j)$$$\Box$

The above is my proof, unlike many books, my proof does not use the property of countable additivity. So I doubt my proof is false. Who can point out where are my mistakes?

  • $\begingroup$ At least somewhere you should remark that $m(A_j)$ is an increasing sequence, and so the limit exists and is less than or equal to any common bound to all terms... $\endgroup$ – Arturo Magidin Apr 22 '12 at 1:55
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    $\begingroup$ First of all, you have $\lim_{N \to \infty} m(A_j)$ on the right-hand side in part of the second step. Should your $N$ be a $j$? Second of all, how do you get from $\lim_{N\to\infty} m(\cup_{j=1}^N A_j)$ to $m(\cup_{j=1}^\infty A_j)$ when taking the limit at the very end? $\endgroup$ – cardinal Apr 22 '12 at 2:00
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    $\begingroup$ In the second half, you are trying to argue that $$\lim_{N\to\infty}m(\cup_{j=1}^NA_j) = m(\lim_{N\to\infty}\cup_{j=1}^NA_j) = m(\cup_{j=1}^{\infty}A_j)$$(in order to "take the limit"). But this is precisely what you are trying to prove. $\endgroup$ – Arturo Magidin Apr 22 '12 at 2:04
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    $\begingroup$ Hint: Write the infinite union as a disjoint union, and use countable additivity. $\endgroup$ – Mike B Apr 22 '12 at 2:35
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    $\begingroup$ @ArturoMagidin: No, I knew you weren't. :) I just was wondering if my statement was a little too obtuse/Socratic. I've noticed the synchronization of how comments get posted can be strange at times. $\endgroup$ – cardinal Apr 22 '12 at 2:49

$m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit, we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$.

The above is valid because if $b_j\le b$ for $j=1,2,3,\ldots$ then $\lim_j b_j \le b$.

$m(\bigcup_{j=1}^{N}A_j)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$.

The above is not valid. You claim to have "taken the limit" of $m\left( \bigcup_{j=1}^N A_j \right)$. You cannot take a limit merely by putting $\infty$ wherever you see $N$. The question is: how do you know that $$ \lim_{N\to\infty} m\left( \bigcup_{j=1}^N A_j \right) = m\left( \bigcup_{j=1}^\infty A_j \right) \text{ ???} $$

First, notice that $\bigcup_{j=1}^\infty A_j$ is not defined as a limit as $N\to\infty$ of $\bigcup_{j=1}^N A_j$. Rather, it is defined by saying $x\in \bigcup_{j=1}^\infty A_j$ if and only if $\exists j\in\{1,2,3,\ldots\}\quad x\in A_j$. And if it were defined as a limit, there would still be the question of continuity of the function $m$. How would you prove that?

Here's a different way: \begin{align} m\left( \bigcup_{j=1}^\infty A_j \right) & = m \left( \bigcup_{j=1}^\infty \left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right)\right) & & \text{(Think about why this is true.)} \\[10pt] & = \sum_{j=1}^\infty m\left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) & & \text{by countable additivity of $m$} \\[10pt] & = \lim_{N\to\infty} \sum_{j=1}^N m\left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) \\[10pt] & = \lim_{N\to\infty} m\left( \bigcup_{j=1}^N A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) & & \text{by finite additivity of $m$} \\[10pt] & = \lim_{N\to\infty} m(A_N). \end{align}

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Of course, as it is mentioned in the comments you were wrong in assuming $$\lim_{N\to\infty}m\left(\bigcup_{j=1}^NA_j\right)=m\left(\bigcup_{j=1}^{\infty}A_j\right)$$ However the proof can be done without using the countably additivity property. Suppose $m$ is a measure defined on $\mathcal{F}$, the field generating the $\sigma$-field $\mathcal{A}$, then for a general $A\in\mathcal{A}$, the outer measure is defined as $$m^{\ast}(A)=\inf\{m(B):B\in\mathcal{F}_{\sigma},B\supset A\}$$ We can show that $m^{\ast}$ is monotone and strongly subadditive. Now let $A_n\uparrow A$.
Then as $A_n\subset A$, so $m^{\ast}(A_n)\leq m^{\ast}(A)$. So $$\lim_{n\to\infty}m^{\ast}(A_n)\leq m^{\ast}(A)$$. Now obviously we can find $B_n\in\mathcal{F}_{\sigma}$ such that $$m(B_n)\leq m^{\ast}(A_n)+\frac{\epsilon}{2^n}$$ by the definition of $m^{\ast}$. Let $$C_n=\bigcup_{k=1}^nB_k$$. Then $A_n\subset B_n\subset C_n$ and hence $\lim A_n\subset\lim C_n$ or $m^{\ast}(A)\leq m^{\ast}(\lim C_n)$. Now $\lim C_n\in \mathcal{F}_{\sigma}$ so $m^{\ast}(\lim C_n)=m(\lim C_n)$ and as $m$ is continuous from below, so $m(\lim C_n)=\lim m(C_n)$. Now $m(C_{n+1})+m(C_n\cap B_{n+1})=m(C_n)+m(B_{n+1})$. Now note that $A_n\subset C_n$ and $A_n\subset A_{n+1}\subset B_{n+1}$. So $A_n\subset C_n\cap B_{n+1}$, so $m^{\ast}(A_n)\leq m^{\ast}(C_n\cap B_{n+1})=m(C_n\cap B_{n+1})$. Thus our claim is proved. So we have $m^{\ast}(A)\leq \lim m(C_n)\leq \lim m^{\ast}(A_n)+\epsilon$. So $$m^{\ast}(A_n)\uparrow m^{\ast}(A).$$

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