Please how can I justify that these two definitions can be the same thing or let me say how can I compare the following two definitions.
If $g$ is a right continuous, increasing step function on $\mathbb{R}$ with countable discontinuous points $x_1,x_2,\cdots,x_n$. Then for any function $f$, we define the integral $$\int_{\mathbb{R} }f dg =\sum_{k=1}^{n} f(x_k)\Delta g(x_k)$$where $\Delta g(x_n) = g(x_n^+)-g(x_n^-)> 0$.
Let $(X, \mathcal{H}, \mu)$ be a measure space and let $f: X \rightarrow [0, \infty]$ be a measurable function. We define the Lebesgue integral of $f$ over $X$ as $$\int_{X}f d \mu= \sup \left\{ \int_{X}g d \mu: 0 \le g \le f, g \,\, \text{is simple} \right\}.$$
I would be glad if a proof or correspondence can be given. Thanks for your help.
The second definition contains the first:
Let $g:\mathbb{R}\rightarrow\mathbb{R}$ as in 1.. With the classical extensions theorems, one can construct a measure $\mu_{g}$ over the borelians such that $\mu_{g}(]a,b])=g(b)-g(a)$. It can be shown that $\mu_{g}(A)=\sum_{x_{n}\in A}\Delta g(x_{n})$.
If $f=\sum_{i=1}^{m}a_{i}1_{A_{i}}$ is measurable (and the $A_{i}$ disjoint) then $$\int fd\mu_{g}=\sum_{i=1}^{m}a_{i}\mu_{g}(A_{i})=\sum_{i=1}^{m}a_{i}\sum_{x_{n}\in A_{i}}\Delta g(x_{n})=\sum_{i=1}^{m}\sum_{x_{n}\in A_{i}}f(x_{n})\Delta g(x_{n})=\sum_{x_n}f(x_{n})\Delta g(x_{n})$$
If$f$ is positive and measurable, then there exist simple functions $f_{k}$ such that $f_{k}(x)\rightarrow f(x)$ monotonously. We deduce, by the monotone convergence theorem and the previous result, that $\int fd\mu_{g}=\sum_{n}f(x_{n})\Delta g(x_{n})$