The following statement comes in handy when dealing with an exercise on Kanamori's book, and since I have little experience with random reals I decided to ask here whether my solution is correct or not.
Q. Assume that $a_1, \dots, a_r, k$ are natural numbers and $x$ is a random real over $V$. Show that there exists $b \in \omega$ such that $b$ is distinct from all the $a_i's$ and that $x(b) = k$.
My attempt:
So I thought about density arguments but I didn't encounter any good dense sets so I decided to use this characterization of random reals: $x$ is random over $V$ iff $x$ is not in any $A_c$ where $c \in V$ is a $G_\delta$ code for a null set. And the main idea is to construct definable $G_\delta$ sets for each $n$ such that $x \not \in A_c$ forces $x$ to take $k$ as a value somewhere above $n$. Since $n$ is arbitrary we can then take it to be $a_r + 1$. Also since the codes will be definable, they will be in $V$.
We solve for $n = 0$, because the other cases will be clear after the proof for $n=0$. For each $i \in \omega$ let $A_i$ be the set of all reals $y$ such that $y(i) \neq k$. Then $$A_i = \bigcup\{O_{s_j} \mid j\in \omega \text{ and } |s_j| = i+1 \text{ and } s_j(i)\neq k\},$$
where the $s_j$'s are a fixed recursive enumeration of $^{\lt\omega}\omega$ and $O_{s_j} = \{ x \in {^\omega\omega} : s_j \subset x \}$. So every $A_i$ is open and we let $c$ be the real coding the intersection of these nicely definable open sets. Now $A_c = \bigcap_{i=0}^{\infty} A_i$. And if we show that $A_c$ is null and then we would get to the desired result as discussed above.
We have that $m_L(A_c) = \inf\{m_L(\bigcap_{i=0}^{l} A_i): l \in \omega \}$, So we let $B_l = m_L(\bigcap_{i=0}^{l} A_i)$. With a usage of the definition of measure and a combinatorial argument we can see that,
$$ B_l = \sum\limits_{i_0 \neq k}\sum\limits_{i_1 \neq k}\dots\sum\limits_{i_l \neq k}(2^{-(i_0 + 1)} \times 2^{-(i_1 + 1)} \dots \times 2^{-(i_l + 1)}).$$
With the help of one of my friends who actually knows how to use the inclusion-exclusion principle(!) we came to this:
$$1 - B_l = \sum_{j = 1}^{l+1}((-1)^{j-1} \times \binom{l+1}{j} \times (2^{-(k+1)})^{j})$$
Also using the binomial theorem we get(again with my friends help!):
$$1 - B_l = -[(1 - 2^{-(k+1)})^{l+1} - 1].$$ Now if we let $l \rightarrow \infty$ then $1 - B_l \rightarrow 1$ and $B_l \rightarrow 0$, which insures that $A_c$ is null. For the case $n > 0$ we can take the corresponding $G_\delta$ set to be $A_c = \bigcap_{i=n}^{\infty} A_i$, which through the argument above will also serve our purpose.
I would be really glad if somebody could point out any potential flaws. Also I would appreciate any reasonably shorter proof of this fact. Sorry for the long post.