Return to Answer

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check$^*$ that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not care about talking about the Borel $\sigma-$algebra on discrete spaces but rather directly declares all subsets of $\Omega$ to be measurable.

$^*$ Let's prove this. In a metric space, a set $A$ is open if for every $x\in A$ there exists an $\epsilon >0$ such that all points in the $\epsilon-$ball around $x$ are also in $A$.

In our example, take $A = \{x\}$ for an arbitrary $x \in \Omega$ and fix an $\epsilon < 1$, say $\epsilon = 1/2$. Then, $x$ is the only point in the open $1/2-$ball around $x$ (recall, the metric is 1 or 0), and $x\in A$ by definition so we conclude $A$ is open. That is, any point is an open set.

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check$^*$ that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not bother with the Borel $\sigma-$algebra on discrete spaces, but instead directly declares all subsets of $\Omega$ to be measurable.

$^*$ Let's prove this. In a metric space, a set $A$ is open if for every $x\in A$ there exists an $\epsilon >0$ such that all points in the $\epsilon-$ball around $x$ are also in $A$.

In our example, take $A = \{x\}$ for an arbitrary $x \in \Omega$ and fix an $\epsilon < 1$, say $\epsilon = 1/2$. Then, $x$ is the only point in the open $1/2-$ball around $x$ (recall, the metric is 1 or 0), and $x\in A$ by definition so we conclude $A$ is open. That is, any point is an open set.

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not care about talking about the Borel $\sigma-$algebra on discrete spaces but rather directly declares all subsets of $\Omega$ to be measurable.

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check$^*$ that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not care about talking about the Borel $\sigma-$algebra on discrete spaces but rather directly declares all subsets of $\Omega$ to be measurable.

$^*$ Let's prove this. In a metric space, a set $A$ is open if for every $x\in A$ there exists an $\epsilon >0$ such that all points in the $\epsilon-$ball around $x$ are also in $A$.

In our example, take $A = \{x\}$ for an arbitrary $x \in \Omega$ and fix an $\epsilon < 1$, say $\epsilon = 1/2$. Then, $x$ is the only point in the open $1/2-$ball around $x$ (recall, the metric is 1 or 0), and $x\in A$ by definition so we conclude $A$ is open. That is, any point is an open set.

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the integers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not care about talking about the Borel $\sigma-$algebra on discrete spaces but rather directly declares all subsets of $\Omega$ to be measurable.

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not care about talking about the Borel $\sigma-$algebra on discrete spaces but rather directly declares all subsets of $\Omega$ to be measurable.