I am currently reading the book "Frank Jones : Lebesgue Integeartion on Euclidean Spaces". The writing is not completely rigorous. for example measure can be $\infty$ but he doesn't tell what to do in such cases.

what does the following statement mean when one of $\lambda(A_i) = \infty.$ (I mean how does infinite series defined when one of them is $\infty$)

$$\lambda\Big( \bigcup_{i=1}^{\infty} A_i \Big) = \sum_{i=1}^{\infty} \lambda(A_i).$$

It is particularly confusing when argument depends on measure of set not being $\infty$

let $A_1 \subset A_2 \subset A_3 \dots$ then $\displaystyle \lambda \Big( \bigcup_{n=1}^{\infty} A_n \Big) = \lim_{N \rightarrow \infty} \lambda(A_N)$

we can write union of $A_i$ as disjoint union as follows $ \displaystyle\bigcup_{n=1}^{\infty} A_n = A_{1} \cup \bigcup_{n=1}^{\infty} (A_{n+1} \sim A_{n}) $ then we have that

$\displaystyle \lambda(\bigcup_{n=1}^{\infty} A_i) = \lambda(A_{1}) + \sum_{n=1}^{\infty} \lambda(A_{n+1} \sim A_{n}) $ he then says that

$ = \displaystyle \lim_{N \rightarrow \infty} \Big[ \lambda(A_{1}) + \sum_{n=1}^{N} \lambda(A_{n+1} \sim A_{n}) \Big]$

$ = \displaystyle \lim_{N \rightarrow \infty} \lambda \Big( A_{1} \cup \bigcup_{n=1}^{N} (A_{n+1} \sim A_{n}) \Big)$

$ = \displaystyle \lim_{N \rightarrow \infty} \lambda(A_N)$

then following is the excercise

Let $A_1 \supset A_2 \supset A_3 \dots$ if $ \lambda(A_1) < \infty$ then we have that $\displaystyle \lambda \Big( \bigcap_{n=1}^{\infty} A_n \Big) = \lim_{n \rightarrow \infty} \lambda(A_i))$

now we know that $\displaystyle A_{K} = \Big( \bigcap_{n=1}^{\infty} A_n \Big) \cup \Big(\bigcup_{n=k}^{\infty}(A_{n} \sim A_{n+1}) \Big) $

so we have that $\lambda(A_{K}) = \lambda(A) + \sum_{n=k}^{\infty} \lambda(A_{n} \sim A_{n+1})$

and $\displaystyle \lim_{K \rightarrow \infty} A_{K} = \lim_{K \rightarrow \infty} \Big[ \lambda(A) + \sum_{n=k}^{\infty} \lambda(A_{n} \sim A_{n+1}) \Big] = \lambda(A)$

But I didn't use the condition that $\lambda(A_1) < \infty.$ ?? what is wrong with the arugment. and I am really confused handling with limit of sequence of numbers which contains $\infty.$ Book doesn't talk about what it means for limit when sequence contains $\infty$ or handle it separately.

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    $\begingroup$ "what does the following statement mean when one of $\lambda(A_i)=\infty$?" It means $\infty=\infty$. $\endgroup$ Commented Jul 29, 2020 at 13:02
  • $\begingroup$ what is limit of following sequence $(0,\infty,0,0,0,0,\dots )$ is it $0$?? $\endgroup$
    – manifold
    Commented Jul 29, 2020 at 13:05
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    $\begingroup$ See page 25: "the measure of $A$ will be a nonnegative real number or $\infty$." And see page 29-30. $\endgroup$ Commented Jul 29, 2020 at 13:07
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    $\begingroup$ If you are in page 44-on, you are working with sets having finite outer measure, i.e. such that $\lambda^*(A) < \infty$ $\endgroup$ Commented Jul 29, 2020 at 13:16
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    $\begingroup$ Presumably, as is explicitly said in some other text, if we have an unbounded family of nonnegative numbers $\{ a_i \}$ we have to set $\Sigma a_i = \infty$. $\endgroup$ Commented Jul 29, 2020 at 13:26

1 Answer 1


For your last example, the assumption that $\lambda(A_j)<\infty$ for some $j$ is crucial.

Let $\lambda$ be Lebesgue measure and $A_k$ be the interval $(k,\infty)$. Then each $\lambda(A_k)=\infty$ but $\bigcap_k A_k=\emptyset$. The usual error here is to assume that "$\infty-\infty$" equals $0$.


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