Given a measure space / probability space $(\Omega, \mathcal{F}, P)$ - one might think about $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \lambda)$, where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$ - we define the first moment (aka. expected value?) for a stocastic variable, X, as $$\text{E\{X\}} =\int_\Omega x \space\text{d}XP(x).$$
Similary one would define the k'th moment as $$\text{E}\{X^k\}=\int_\Omega x^k \space \text{d} XP(x).$$
However right after this introduction I was given the following example - both by my professor and my book (combining the definition above with another result. I do not know which result):
Let $X$ be a real-valued stochastic variable which has density $f$ with respect to the Lebesgue measure. We see that $$\text{E}\{X^k\}=\int_{\mathbb{R}}x^kf(x) \space \text{d}x.$$
My questions are: Why do we now multiply by $f$ under the integration? It seems to me that both the definition and the example of the k'th moment are valid, but they are not identical, so what is going on? And if $k=1$ (meaning that we are looking at the first moment), which of the cases is then the expected value?
You can have a look at, Properties section of https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem
Edit: (I hope I didn't misunderstand your question) Let $X: \Omega \to \mathbb{R}$ be a measurable function (in our case the random variable) ($ \forall ;A \in \mathcal{B}(\mathbb{R}), \ X^{-1}(A)\in \mathcal{F}$). Then, we get a probability measure $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{P}_X)$, $\mathcal{P}_X(A) = \mathcal{P}(X^{-1}(A))$. Then, $$ \int_{\Omega}X^k(\omega)d\mathcal{P} = \int_{\mathbb{R}}x^kd\mathcal{P}_X $$ which can be shown with the definition of the Lebesgue integral. Then using the Radon-Nikodym theorem, there exists a probability density function $f = \frac{d\mathcal{P}_X}{d\mu}$, such that for any integrable $g$, $$\int_{\mathbb{R}}g \ d\mathcal{P}_X = \int_{\mathbb{R}}g f\ d\mu$$ Substituting $x^k$ for $g$ we get the result.