Given a probability space $\left([0,5], \mathcal{B}([0,5]), \lambda_{1} / 5\right)$ and the random variable $X(w)=I_{[0,1]}(\omega)-2 I_{[2,4]}(\omega)$ I need to determine $E(X)$.
My understanding is that I need to evaluate $X(w)$ for $w$ in $[0, 5]$, and then $E(X)$ is simply $\sum_{w} X(w) P(w)$.
Does $\lambda_1/5$ mean that $\mathbb{P}(w)=0.2$ for all $w$?
On
I add some explanations in plain English. Assume $\lambda_1$ is a Lebesgue measure e.g., for any interval $(a,b)\subset[0,5]$ we have $\lambda_1((a,b))=b-a.$
Then this is a continuous random variable which is uniformly distributed on $[0,5].$ There is no probability mass for any arbitrary value of a continuous random variable.
In this manner, expectation of an indicator random variable is the measure of the set which the indicator random variable belongs to divided by the length of [0,5]. The rest follows the answer by Ivo Trek.
No. It means that for every $E\in \mathcal{B}[0,5]$ you have $\Bbb P(E) = \lambda_1(E)/5$. In particular, $\Bbb P(\omega)=0$ for all $\omega\in [0,5]$ since points have zero Lebesgue measure. In general, we have
$$\Bbb E[X]=\int_\Omega X(\omega)\,{\rm d}\Bbb P(\omega),$$ right? In your case we get $$\Bbb E[X]=\int_{[0,5]} I_{[0,1]}(\omega) -2I_{[2,4]}(\omega)\,{\rm d}\Bbb P(\omega) = \Bbb P([0,1])-2\Bbb P([2,4]) =\frac{1}{5}-2\frac{2}{5}=-\frac{3}{5}.$$