Suppose that $\Omega$ is a set, $(\Omega, \mathscr{G})$ is a measure space, and $Z: \Omega \to \mathbb{R}$ is a given mapping. Then Z is $\mathscr{G}$ measurable iff $$Z =\displaystyle\sum_{i=1}^\infty \lambda_i I_{A_i} \tag{1}$$ for some {$\lambda$} $\subset \mathbb{R}$ and {$A_i$} $\subset \mathscr{G}$. From(1) deduce the First Borel-Cantelli Lemma: if $\displaystyle\sum_{i=1}^{\infty} P(A_i) < \infty $ then $P(\lim sup A_n) =0$

Solution:= The "if" part follows directly from the observation that the mapping given by (1) is the pointwise limit as $n \to \infty $ of the $\mathscr{G}$ measurable mappings $\displaystyle\sum_{i=1}^n \lambda_i I_{A_i}.$

For the opposite direction, first suppose that Z is nonnegative . Define $$\begin{align*} Z_1 := I_{Z \geq 1},\\ & S_n :=\displaystyle\sum_{i=1}^n \displaystyle\frac{1}{i}Z_i, Z_{n+1} := I_{Z-\displaystyle\frac{1}{n+1}\geq S_n}, n\geq 1 \end{align*}$$

We claim that $Z = \lim S_n = \displaystyle\sum_{i=1}^\infty \displaystyle\frac{1}{i} Z_i $ (Note that $\lim S_n$ exists).

Using induction we first show that $S_n \leq Z, \forall n \in \mathbb{N}$ Clearly $S_1 = I_{Z \geq 1} \leq Z.$ Suppose that the claim holds for some $m \in \mathbb{N}$ and denote {$ Z -\displaystyle\frac{1}{m+1} \geq S_m$} by A. Then on A, we have $Z \geq \displaystyle\frac{1}{m+1} + S_m $ while, by the inductive hypothesis, on $A^c$ we have $Z \geq S_m$.Thus $Z \geq \displaystyle\frac{1}{m+1} I_A +S_m= S_{m+1}$ which completes the proof by induction.

It remains to show that $\lim S_n \geq Z$. Fix an arbitrary $\omega \in \Omega$. If $Z ( \omega) = \infty$, we have $Z_n( \omega) = 1 \forall n \in \mathbb{N} $ and the result follows. If $Z(\omega) < \infty ,$ suppose that on the contrary $Z(\omega) -\lim S_n > 0 $ and find $ k \in \mathbb{N} $ such that $\displaystyle\frac{1}{k} < Z(\omega) -\lim S_n$. Then, since $ S_n \leq \lim S_n,$ we have $Z(\omega)-\displaystyle\frac{1}{n+1}\geq S_n(\omega), \forall n \geq k,$ and so $Z_n(\omega) =1 \forall n \geq k.$ But then $\lim S_n =\infty $ which is a contradiction.

In the general case we write $Z = Z^+ - Z^- $ and apply the above result to both parts (since$ Z^+ $ and $Z^-$ live on disjoint sets there is no additional convergence issues in merging the two series.)

The First Borel-Cantelli Lemma is essentially the statement that if $Z = \displaystyle\sum_{i=1}^\infty I_{A_i}$ is integrable, then $Z < \infty $ almost surely.

Would any member of math stack exchange explain this solution to the given problem with all the details and how,where,when and for what purposes and in which situations this results can be used in the real world scenario?🤔🤔🤔

For example, suppose I want to make risk assessment for flood insurance where I as an insurer need to estimate the probability of a certain amount of rainfall occurring. In this situation, How can I use this result of First Borel-Cantelli Lemma?🤔🤔🤔


1 Answer 1


The First Borel-Cantelli Lemma can be applied in risk assessment for flood insurance in the following way:

The lemma states that if you have an infinite sequence of events, and the sum of their probabilities is finite, then the probability that any of these events will occur infinitely often is zero.

In the context of flood insurance, consider each "event" as a flood occurring in a specific area. If the probability of a flood occurring in each year is independent of the previous years, and the sum of these probabilities over an infinite time horizon is finite, then according to the First Borel-Cantelli Lemma, the probability that floods will occur infinitely often in this area is zero.

This can help insurance companies in their risk assessment. If the conditions of the lemma are met, they can conclude that the risk of having to pay out for flood damage every year is zero. This can influence the pricing of their insurance policies and their decision on whether to offer insurance in that area.

However, it's important to note that in reality, events like floods are often not independent and their probabilities may not be easy to estimate accurately. Therefore, while the Borel-Cantelli Lemma provides a theoretical framework, its application in real-world scenarios like flood insurance should be done with caution and in conjunction with other risk assessment tools.

The problem is about measure theory, which is an area of mathematics concerned with measuring the “size” of subsets of a certain set. A measure is a function that assigns a non-negative number to each subset of a given set, satisfying some properties such as countable additivity. A measure space is a triple consisting of a set, a collection of subsets called a sigma-algebra, and a measure defined on that sigma-algebra. A mapping from a measure space to the real numbers is called measurable if it preserves the structure of the sigma-algebra, that is, if the inverse image of any measurable set is also measurable.

The problem asks to prove that a mapping Z is measurable if and only if it can be written as an infinite sum of indicator functions of measurable sets, where an indicator function is a function that takes the value 1 on a given set and 0 elsewhere. The problem also asks to deduce from this result the First Borel-Cantelli Lemma, which is a statement about the probability of the limit superior of a sequence of events, where the probability is a special kind of measure that assigns values between 0 and 1 to each event.

The solution proceeds as follows:

  • The "if" part of the problem is easy to show, because if Z can be written as an infinite sum of indicator functions of measurable sets, then for any measurable set B, the inverse image of B under Z is the union of the inverse images of B under each indicator function, and since the union of measurable sets is measurable, the inverse image of B under Z is also measurable.
  • The "only if" part of the problem is more involved, and it requires to consider two cases: when Z is non-negative and when Z is general. The idea is to construct a sequence of measurable functions that converges pointwise to Z, and then use the fact that the limit of measurable functions is measurable.
  • In the case when Z is non-negative, the solution defines a sequence of functions $Z_n$ as follows: $Z_1$ is the indicator function of the set where Z is greater than or equal to 1, and for n greater than 1, $Z_n$ is the indicator function of the set where Z minus $\frac{1}{n+1}$ is greater than or equal to the sum of the previous functions. The solution also defines another sequence of functions $S_n$ as the partial sums of the products of $Z_n$ and $\frac{1}{n}$. The solution claims that Z is equal to the limit of $S_n$, which is also equal to the infinite sum of the products of $Z_n$ and $\frac{1}{n}$.
  • To prove this claim, the solution first shows by induction that $S_n$ is less than or equal to Z for all n, and then shows that the limit of $S_n$ is greater than or equal to Z for all $\omega$ in the measure space. The latter part is done by contradiction, assuming that there exists an $\omega$ where $Z(\omega)$ minus the limit of $S_n$ is positive, and then finding a contradiction by showing that $Z_n(\omega)=1$ for all sufficiently large n, which implies that the limit of $S_n$ is infinite.
  • In the case when Z is general, the solution writes Z as the difference of two non-negative functions $Z^+$ and $Z^-$, and applies the previous case to each of them separately. Since $Z^+$ and $Z^-$ are zero on disjoint sets, there is no problem in adding their respective series to obtain the series for Z.
  • The First Borel-Cantelli Lemma is a consequence of the previous result, because if Z is the infinite sum of the indicator functions of a sequence of events $A_i$, and if the sum of the probabilities of $A_i$ is finite, then Z is integrable, that is, the measure of Z is finite. This implies that Z is finite almost surely, that is, except on a set of measure zero. But the set where Z is infinite is precisely the limit superior of the sequence of events $A_i$, that is, the set of points that belong to infinitely many $A_i$. Therefore, the probability of the limit superior of $A_i$ is zero, as the lemma states.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .