Finding $\int_{[0,1]}f\,d\mu$

Question

I am trying to solve this task

Consider the measure space $([0,1],\mathcal{B}([0,1]),\mu:=2\lambda_1+3\delta_{1})$ and the measurable function $f(\omega)=2-\omega$. Find $$\int_{[0,1]}f\,d\mu$$

My Solution

$$\int_{a}^{b} f(x) d x=\int_{[a, b]} f d \lambda_{1}$$

We know

Thus

$$\int_{[0,1]} f d \mu=\int_{[0,1]} 2- x\left(2d \lambda_{1}+3 d \delta_{1}\right)=\cdots=6$$

Answer 1

I pressume that $\delta_1$ is the Dirac measure and $\lambda_1$ the Lebesgue measure.

If so,then you are correct.

Indeed:

So $$\int_0^1f(x)d\mu(x)=2\int_0^1(2-x)d\lambda_1(x)+3\int_0^1f(x)d\delta_1(x)$$ $$=2\int_0^1(2-x)dx+3 f(1)=2(2-\frac{1}{2})+3=6$$

Answer 2

Marios answer is correct

Just wanted to note to the OP some general facts. It can be checked for simple functions, and then proven in general, that for $a,b$ non-negative and $\mu = a\lambda + b\delta$ where $\lambda,\delta$ finite on the measurable set $A$, we have $$ \int_{A} f d\mu = a\int_{A} f d\lambda + b\int_{A} f d\delta $$