Let $P$ and $Q$ be two probability measures defined on the same collection of events that, for each $\alpha \in [0, 1]$. Let $P$ and $Q$ be two probability measures defined on the same collection of events that, for each $\alpha \in [0, 1]$, the function $\alpha P+(1-\alpha)Q$ is also a probability measure.
Attempt
I believe that it must be proved that the function $A\rightarrow \alpha P(A)+(1-\alpha)Q(A)$ satisfies the three axioms of probability.
Given a measurable space $(\Omega,F)$ that $P, Q$ are probability measures for, it must hold:

*

*$\forall  A\in \mathcal{F}: \alpha P(A)+(1-\alpha)Q(A)\in [0, 1]$


*$\alpha P(\Omega)+(1-\alpha)Q(\Omega)=1$.
Since $P(\bigcup\limits_{i=1}^{\infty} A_{i})=\sum_{i=1}^{\infty}P(A_i)$ and $Q(\bigcup\limits_{i=1}^{\infty} A_{i})=\sum_{i=1}^{\infty}Q(A_i)$, then

*

*$\alpha P(\bigcup\limits_{i=1}^{\infty} A_{i})+(1-\alpha)Q(\bigcup\limits_{i=1}^{\infty} A_{i})=\sum_{i=1}^{\infty}\alpha P(A_i)+(1-\alpha)Q(A_i)$
 A: Yes.  That is what you need to do, and roughly how you need to do it.  But you should polish it a bit more.
As provided, $1\leqslant\alpha\leqslant 1$, and that $P$ and $Q$ are probability measures over the same event space $(\Omega, \mathcal F)$
This means $P$ and $Q$ obey the three axioms, and we must show that $\alpha P+(1-\alpha)Q$ does too.

Axiom One: Probability of an Event.
For any $A\in\mathcal F$, we have $0\leqslant P(A)\leqslant 1$ and $0\leqslant Q(A)\leqslant 1$.
Therefore also $0\leqslant \alpha P(A) + (1-\alpha)Q(A)\leqslant \alpha+(1-\alpha) = 1$

Axiom Two: Probability of the Outcome Set (Sample Space)
We have $P(\Omega)=1$ and $Q(\Omega)=1$.
Therefore also $\alpha P(\Omega)+(1-\alpha) Q(\Omega)=1$.
[ Likewise $P(\emptyset)=0, Q(\emptyset)=0$ implies $\alpha P(\emptyset)+(1-\alpha) Q(\emptyset)=0$.  This is not always included among the axioms, but...eh.]

Axiom Three: Additivity of Probability for Unions of Disjoint Events.
For any sequence of disjoint events $(A_i)$, we have $P(\bigcup_i A_i)=\sum_i P(A_i)$ and $Q(\bigcup_i A_i)=\sum_i Q(A_i)$
Therefore also $\alpha P(\bigcup_i A_i)+(1-\alpha)Q(\bigcup_i A_i)=\sum_i\left(\alpha P(A_i)+(1-\alpha)Q(A_i)\right)$

So if $(\Omega,\mathcal F, P)$ and $(\Omega, \mathcal F, Q)$ are probability spaces, then $(\Omega,\mathcal F, [\alpha P +(1-\alpha)Q] )$ is also a probability space.
