[proofplan] Fix a threshold $t>0$. The simultaneous deviation event is exactly the union of the individual deviation events because the maximum over the finite index set exceeds $t$ precisely when at least one coordinate exceeds $t$. Since probability measures are finitely subadditive, the probability of this union is bounded by the sum of the individual probabilities. The assumed one-dimensional tail bounds then give the desired estimate. [/proofplan] [step:Rewrite the simultaneous deviation event as a finite union] Fix $t>0$. For each $j\in J$, define the event $A_j\in\mathcal F$ by \begin{align*} A_j:=\{\omega\in\Omega: |Y_j(\omega)|\ge t\}. \end{align*} The measurability of $A_j$ follows because $Y_j$ is a real-valued [random variable](/page/Random%20Variable) and $(-\infty,-t]\cup[t,\infty)\in\mathcal B(\mathbb R)$, with \begin{align*} A_j=Y_j^{-1}\bigl((-\infty,-t]\cup[t,\infty)\bigr). \end{align*} Define the simultaneous deviation event $A\in\mathcal F$ by \begin{align*} A:=\left\{\omega\in\Omega: \max_{j\in J}|Y_j(\omega)|\ge t\right\}. \end{align*} Since $J$ is finite and nonempty, for every $\omega\in\Omega$, \begin{align*} \max_{j\in J}|Y_j(\omega)|\ge t \end{align*} holds if and only if there exists $j\in J$ such that $|Y_j(\omega)|\ge t$. Hence \begin{align*} A=\bigcup_{j\in J}A_j. \end{align*} [guided] Fix a threshold $t>0$. We want to compare the event that at least one of the random variables has a large absolute value with the individual large-deviation events. For each $j\in J$, define \begin{align*} A_j:=\{\omega\in\Omega: |Y_j(\omega)|\ge t\}. \end{align*} This is an event in $\mathcal F$: indeed, since $Y_j:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $(-\infty,-t]\cup[t,\infty)$ is a Borel subset of $\mathbb R$, we have \begin{align*} A_j=Y_j^{-1}\bigl((-\infty,-t]\cup[t,\infty)\bigr)\in\mathcal F. \end{align*} Now define \begin{align*} A:=\left\{\omega\in\Omega: \max_{j\in J}|Y_j(\omega)|\ge t\right\}. \end{align*} Because $J$ is finite and nonempty, the maximum $\max_{j\in J}|Y_j(\omega)|$ is well-defined for every $\omega\in\Omega$. For a fixed outcome $\omega\in\Omega$, the inequality \begin{align*} \max_{j\in J}|Y_j(\omega)|\ge t \end{align*} means exactly that at least one of the finitely many numbers $|Y_j(\omega)|$ is at least $t$. Therefore $\omega\in A$ if and only if $\omega\in A_j$ for some $j\in J$. This proves the identity \begin{align*} A=\bigcup_{j\in J}A_j. \end{align*} [/guided] [/step] [step:Apply finite subadditivity and insert the tail bounds] Using the identity from the previous step and finite subadditivity of the probability measure $\mathbb P$, \begin{align*} \mathbb P(A) =\mathbb P\left(\bigcup_{j\in J}A_j\right) \le \sum_{j\in J}\mathbb P(A_j). \end{align*} By the hypothesis applied to each $j\in J$ at the fixed threshold $t>0$, \begin{align*} \mathbb P(A_j) =\mathbb P(\{\omega\in\Omega: |Y_j(\omega)|\ge t\}) \le p_j(t). \end{align*} Substituting these bounds into the finite subadditivity estimate gives \begin{align*} \mathbb P\left(\left\{\omega\in\Omega: \max_{j\in J}|Y_j(\omega)|\ge t\right\}\right) =\mathbb P(A) \le \sum_{j\in J}p_j(t). \end{align*} Since $t>0$ was arbitrary, the desired inequality holds for every $t>0$. No independence assumption is used. [/step]