In the following, I tried to get cdf of random variable $X$, which has an $\operatorname{Exp}(1)$ distribution, by lebesgue integral.
Probability space $(\Omega, \mathcal{F}, \mathbb{P})$: $\mathbb{P}$ a probability measure on $\Omega$. A random variable $X$ is a deterministic function $X:\Omega\to\mathbb{R}$. Distribution of $X$: determined by how $\mathbb{P}$ assigns probabilities to subsets $\Omega$ and how $X$ maps those to subsets of $\mathbb{R}$.
An example: $\Omega = [0,1]$, and $\mathbb{P}$ is uniform on $\Omega$, i.e., for $0\leq a\leq b\leq 1$: $\mathbb{P}[a,b] = b-a$. Define $X:[0,1]\to\mathbb{R}$ by $X(w) = -\log w$. Under $\mathbb{P}$ the rv $X$ has an $\operatorname{Exp}(1)$ distribution.
First, I partitioned y-axis: $[e^{-\frac{i}{n}},e^{-\frac{i-1}{n}}]$ So the found lebesgue integral was $$ \operatorname*{lim}_{n\to\infty}\left(\sum_{i=1}^{n}{\frac{ix}{n}\left(e^{-((i-1)/n)\,x}-e^{-(i/n)\,x}\right)}\right) = 1 - \frac{1+x}{e^x} $$
It should be $1-e^{-x}$ but I got wrong in somewhere.
Could you explain where i misunderstood and how to get cdf?
As pointed out in the comment, there are more straightforward ways to do this. Even if you want to be extra formal, you can observe that the cdf $F$ of $X$ is, by definition: $$F(x) = P(X\leq x) = \mathbb{P}\bigl(\{\omega\in\Omega : X(\omega)\leq x\}\bigr)$$ and then just write, for $0\leq x\leq 1$: $$\ldots = \mathbb{P}\bigl(\{\omega\in[0,1]:-\log(\omega)\leq x\}\bigr) = \mathbb{P}\bigl(\{\omega\in[0,1]:\omega\geq e^{-x}\}\bigr) = \mathbb{P}[e^{-x},1] = 1-e^{-x} $$
In your attempt to evaluate the Lebesgue integral by summation, it looks like your limit-of-the-sum on the left-hand side corresponds to the correct integral, but the right-hand side is incorrect.