Let $\mu$ be Dirac measure i.e. $ \mu_a(A) =\begin{cases} 0 &\text{for} \ a\not\in A \\1 &\text{for} \ a \in A\end{cases}$ . I have to show, that $ \mu: \mathcal{P}(\mathbb{R}) \ni A \rightarrow \sum_{n \in \mathbb{N}} \mu_{2^{-n}}(A) \in [0,\infty]$ is measure.
So, here I have problem with the second property : $ \mu (\bigcup_{k=1}^{\infty} E_k)=\sum_{k=1}^{\infty}\mu(E_k)$
Now I trying to calculate integrals: $\int_{[0,2015]} k d\mu$ where $k(x)=x$
1.$\int_{[0,2015]} k d\mu= \sum_{n \in \mathbb{N}} k(2^{-n})= \sum_{n \in \mathbb{N}} 2015 (2^{-n})$. Is this correct?
2.$\int_{[0,1] \times {[0,\pi]}} x \cos y d(\mu \times \lambda_1)$
I think that it will be zero, because we have : $\int_{[0,1]}(\int_{[0,\pi]} x \cos y d\mu)d \lambda_1$
If $\{E_k\}_1^\infty$ is a countable set of disjoint sets then $$ \mu(\bigcup_{k=1}^{\infty} E_k) = \sum_{n \in \mathbb N} \mu_{2^{-n}}(\bigcup_{k=1}^{\infty} E_k) = \sum_{n \in \mathbb N} \sum_{k=1}^{\infty} \mu_{2^{-n}}(E_k) = \sum_{k=1}^{\infty} \sum_{n \in \mathbb N} \mu_{2^{-n}}(E_k) = \sum_{k=1}^{\infty} \mu(E_k). $$
Your first integral to determine is $$ \int_{[0,2015]} x \, d\mu(x) = \sum_{n=0}^{2015} 2^{-n}. $$ (Here I have assumed that $0 \in \mathbb N.$)
Your second integral to determine is $$ \int_{[0,1] \times {[0,\pi]}} x \cos y \, d(\mu \times \lambda_1) = \left( \int_{[0,1]} x \, d\mu \right) \left( \int_{[0,\pi]} \cos y \, d\lambda_1 \right) = \left( \int_{[0,1]} x \, d\mu \right) \cdot 0 = 0 $$