Probability of Continuous Random Variable Lying in Singleton Set is Zero/Lemma

Lemma

Let $x$ be a real number.

Then:

$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$

Proof

We first show that:

$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} \subseteq \openint {-\infty} x$

Let:

$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$

Then, for some $k \in \N$ we have:

$\ds t \in \hointl {-\infty } {x - \frac 1 k}$

In particular, we have:

$t \le x - \dfrac 1 k$

Since:

$-\dfrac 1 k < 0$

we certainly have:

$t < x$

and hence:

$t \in \openint {-\infty} x$

So, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} \subseteq \openint {-\infty} x$

from the definition of subset.

We now show that:

$\ds \openint {-\infty} x \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$

Suppose that:

$t \in \openint {-\infty} x$

Then either:

$t \le x - 1$

or:

$x - 1 < t < x$

If:

$t \le x - 1$

then:

$\ds t \in \hointl {-\infty} {x - 1}$

so:

$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$

Now suppose that:

$x - 1 < t < x$

Since:

$x_n \to x$

and $\sequence {x_n}_{n \mathop \in \N}$ is increasing, there exists $N \in \N$ such that:

$\ds t \le x_N$

That is:

$\ds t \le x - \frac 1 N$

so:

$\ds t \in \hointl {-\infty} {x - \frac 1 N}$

giving:

$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$

We therefore have:

$\ds \openint {-\infty} x \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$

So, from the definition of set equality, we have:

$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$

$\blacksquare$