Proof of easy direction of van Lambalgen's theorem from Downey & Hirschfeldt This question regards the "easy" direction of van Lambalgen's theorem as proven on page 258 of Downey & Hirschfeldt's Algorithmic Randomness and Complexity. Specifically, in their proof of the following fact:
Claim. If $A \oplus B$ is $1$-random then $B$ is $1$-random relative to $A$.
Their proof goes verbatim like this:

Proof. Suppose that $B$ is not $1$-random relative to $A$. Let $U_{0}, U_{1}, \ldots$ be a universal oracle Martin-Löf test. Then $B \in \bigcap_{i} U^{A}_{i}$. Let $V_{i} = \{[\sigma \oplus \tau]: \tau \in U^{\sigma}_{2i} \text{ and } |\sigma|=i\}$. The $V_i$ are uniformly $\Sigma^{0}_{1}$ classes and $\mu(V_{i}) \leq 2^{-i}$ because $\mu(U^{X}_{2i})\leq 2^{-2i}$ for all $X$. Thus $V_{0}, V_{1}, \ldots$ is a Solovay test. But $A\oplus B \in V_{i}$ since there are infinitely many initial segments $\sigma \prec A$ and $\tau \prec B$ with $\sigma \oplus \tau$ in some $U_{2i}$. So $A \oplus B$ is not $1$-random.

Question 1: It seems there's a conflict between saying $V_{0}, V_{1}, \ldots$ is a Solovay test and saying $A\oplus B \in V_i$. I guess they must mean that $A \oplus B \in V_{i}$ for infinitely many $i$. Is my guess here wrong?
Question 2: I don't see why $A\oplus B$ must be in any of the $V_{i}$'s. For $A\oplus B$ to be in $V_i$ means that $i$ bits of $A$ as an oracle is sufficient to guarantee that $B$ enters $U^{A\upharpoonright i}_{2i}$. It seems plausible, however that for any $i$, the amount of oracle needed be greater than $i$, even really fast growing. Why must $A\oplus B$ be in infinitely many $V_{i}$'s?
 A: Question 1 : I think your guess is correct.
Question 2 : I agree with you and don't understand either the proof the way it is written. Also when you write $V_i=\{[\sigma \oplus \tau]: \tau \in U_{2i}^\sigma$ and $|\sigma|=i\}$, it seems that $\sigma \oplus \tau$ makes sense only when $\sigma$ and $\tau$ have the same length. But when we keep only the $\tau$ of length $i$, each $V_i$ is then empty, because there is no $\tau$ of length $i$ in $U_{2i}$. I propose the following fix:
let $V_{i, \sigma}=\{[\sigma \oplus \tau]: [\tau] \subseteq U_{i}^\sigma$ and $|\tau|=|\sigma|\}$ and let $V_i=\bigcup_{\sigma \in 2^{<\omega}} V_{i,\sigma}$. Note that we also have $V_i=\bigcup_n \bigcup_{\sigma \in F_n} V_{i, \sigma}$, where $F_n$ is the set of strings of length $n$.
For a given $n$ and a given $\sigma$ of length $n$ we have $\mu(V_{i, \sigma}) \leq 2^{-i}2^{-n}$. Then for any $n$ we have $\mu(\bigcup_{\sigma \in F_n} V_{i, \sigma}) \leq 2^{-i}$. We also easily see that for $\sigma_1 \preceq \sigma_2$ we have $V_{i, \sigma_1} \subseteq V_{i, \sigma_2}$. Then for $n_1 \leq n_2$ we have $\bigcup_{\sigma \in F_{n_1}} V_{i, \sigma} \subseteq \bigcup_{\sigma \in F_{n_2}} V_{i, \sigma}$ and then $\mu(\bigcup_n \bigcup_{\sigma \in F_n} V_{i, \sigma})=\sup_{n}\mu(\bigcup_{\sigma \in F_n} V_{i, \sigma}) \leq 2^{-i}$.
This the $\{V_i\}_{i \in \omega}$ are a Martin-Löf test. As for each i we have $B \in U_i^A$, we also have for each $i$ a prefix $\sigma$ of $A$ and a prefix $\tau$ of $B$ with $\sigma=\tau$ and so that $[\tau] \subseteq U_i^\sigma$. Then for each $i$ we have $A \oplus B \in V_i$ and thus $A \oplus B$ is not Martin-Löf random.
I hope this is correct.
