Take $X_N = XI(0<X \leq N)$ and $E[X] = lim_{n \rightarrow \infty} \sum_{i=1}^n {i-1 \over \sqrt n} P({i-1 \over \sqrt n} < X \leq {i \over \sqrt n})$
How is the following equality true? $E[X_N] = \int_0^N x \space dF(x) = \int_0^N (F(N) - F(x)) \space dx$
$E[X_N] = lim_{n \rightarrow \infty} \sum_{i=1}^n {i-1 \over \sqrt n}I(0 < N \leq N) P({i-1 \over \sqrt n} < X \leq {i \over \sqrt n})$
$= lim_{n \rightarrow \infty} \sum_{i=1}^n {i-1 \over \sqrt n}I(0 < X \leq N) (F({i \over \sqrt n})- F({i-1 \over \sqrt n}))$
$= lim_{n \rightarrow \infty} \sum_{i=1}^n {i-1 \over \sqrt n}I(0 < X \leq N) F({i \over \sqrt n}) -lim_{n \rightarrow \infty} \sum_{i=1}^n {i-1 \over \sqrt n}I(0 < X \leq N) F({i-1 \over \sqrt n})$
From here I am confused as to how one proceeds to bring the limit of sum to an integral. Could I have some hint? Moreover can I have some intuition behind the expression $E[X_N] = \int_0^N x \space dF(x)$ It is to me quite an abstract expression.
If $F$ is the CDF of $X$, then using integration by parts, \begin{align} \mathsf{E}X_N&=\int_{-\infty}^\infty x1\{0<x\le N\}dF(x)=\int_{0}^N xdF(x) \\ &=xF(x)\mid_0^N-\int_0^N F(x)dx=NF(N)-\int_0^N F(x)dx. \end{align}