Two definitions of Lebesgue integration

I’ll paraphrase the slicing argument given in Fremlin’s Measure theory, Volume 2, 252O, page 220. Since I’m not going to base a theory of integration on this formula, I think it’s legitimate to appeal to (very) basic Lebesgue theory (no Fubini, only monotone convergence) to prove the desired identity. I leave it to you to deal with the necessary but easy modifications if you allow extended real-valued functions or partially defined functions:

Let $(X,\Sigma,\mu)$ be a measure space and let $f\colon X \to [0,\infty)$ be a measurable function. Then for the usual Lebesgue integral $\int_X f\,d\mu$ of $f$ over $(X,\Sigma,\mu)$ we have the identity
$$\int_X f\,d\mu = \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt$$
where the second integral is a Lebesgue integral of a non-increasing and non-negative function, hence the integrand is continuous up to a countable subset of $[0,\infty)$, and thus its Riemann integral exists in the extended sense and the Riemann and Lebesgue integrals coincide for it.

To prove this, set $E^{n}_k = \left\{x \in X\,:\,f(x) \gt \frac{k}{2^{n}}\right\}$ and put
$$g_n = \frac{1}{2^n} \sum_{k=1}^{4^n} [E_{k}^n]$$
where $[A]$ denotes the characteristic function of $A$. Note that we have $0 \leq f(x) - g_n(x) \lt 2^{-n}$ whenever $0 \leq f(x) \lt 2^n$, so that $g_n(x) \to f(x)$ for every $x \in X$. Also, $g_n(x) \leq g_{n+1}(x)$ for all $x$, so the sequence $(g_n)_{n \in \mathbb{N}}$ increases to $f$ pointwise everywhere. By the monotone convergence theorem we conclude that
$$\tag{1} \int_X f\,d\mu = \lim\limits_{n\to\infty} \int_X g_n\,d\mu.$$
Moreover, for every $t \in [0,\infty)$, we have
$$\{x\in X\,:\,f(x) \gt t\} = \bigcup_{n \in \mathbb{N}} \{x \in X\,:\,g_n(x) \gt t\},$$
so, using the notation in the question, we have for all $t \in [0,\infty)$ that
$$F_f(t) = \mu\{x\in X\,:\,f(x) \gt t\} = \lim\limits_{n\to\infty} \mu\{x \in X\,:\,g_n(x) \gt t\} = \lim\limits_{n\to\infty} F_{g_n}(t)$$
where the convergence is monotone in $n$. Again by the monotone convergence theorem we get
$$\tag{2} \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt = \lim\limits_{n\to\infty} \int_{0}^\infty \mu\{x \in X\,:\,g_n(x) \gt t\}\,dt.$$
Recalling the definitions of $E_{k}^n$ and $g_n$ we see that
$$\mu E_{k}^n = \begin{cases} \mu\{x\in X\,:\,g_n(x) \gt t\} & \text{if }1 \leq k \leq 4^n \text{ and } \frac{k-1}{2^n} \leq t \lt \frac{k}{2^n} \\ 0, & \text{otherwise,} \end{cases}$$
so that
$$\tag{3} \int_{0}^\infty \mu\{x\in X\,:\,g_n(x) \gt t\}\,dt = \sum_{k=1}^{4^n} \frac{1}{2^n}\mu E_{k}^n = \int_{X} g_n\,d\mu.$$
Combining $(1)$ and $(2)$ using $(3)$ we get the desired formula
$$\int_X f\,d\mu = \int_{0}^\infty \mu\{x\in X\,:\,f(x) \gt t\}\,dt.$$