Let $(\mu_n)_{n\geq 1}$ be a sequence of probability measures with $\mu_n(\Omega)=1$ in a measureable space $(\Omega, \mathcal{F})$. Show that the function
$$\lambda:\mathcal{F}\longrightarrow[0,\infty], E\rightarrow \sum_{n=1}^n \frac{\mu_n(E)}{2^n}$$ is a probability measure on $(\Omega, \mathcal{F})$.
Professor didn't cover probability measures, he just told us to look at the Wikipedia page on probability measure.
Just verify the axioms of the probability:
$$\lambda(\Omega)=\sum_{n=1}^\infty\frac{\mu_n(\Omega)}{2^n}=\sum_{n=1}^\infty\frac{1}{2^n}=1, \lambda(\emptyset)=\sum_{n=1}^\infty\frac{\mu_n(\emptyset)}{2^n}=0$$ and: $0\le\lambda(E)\le 1$, which is a result of the obvious $0\le \mu_n(E)\le 1$:
$$0\le\sum_{n=1}^\infty\frac{\mu_n(E)}{2^n}\le \sum_{n=1}^\infty\frac{1}{2^n}=1$$
and: $$\lambda(\bigcup_{i=1}^\infty E_i)=\sum_{n=1}^\infty\frac{\mu_n(\bigcup_{i=1}^\infty E_i)}{2^n}=\sum_{n=1}^\infty\sum_{i=1}^\infty \frac{\mu_n(E_i)}{2^n}=\sum_{i=1}^\infty\sum_{n=1}^\infty \frac{\mu_n(E_i)}{2^n}=\sum_{i=1}^\infty\lambda(E_i)$$ with the mention that the sums could be interchanged because the series involved are absolutely convergent, having positive terms.