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:


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