calculus - Understanding $mathbb{E}((X-c) mathbb{1}_{{X > c}}) = int_c^{+infty}mathrm P(Xgeqslant t),mathrm dt$

For example, in this MSE post, it is claimed that the following holds:

$$\mathrm E(|X-c|)=\int_{-\infty}^c\mathrm P(X\leqslant t)\,\mathrm dt+\int_c^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt$$

The modulus part is clear, so it suffices to stick with the first part, namely:

$$\mathbb{E}((X-c) \mathbb{1}_{\{X > c\}}) = \int_c^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt$$

Why does this hold? What I'm seeing is that:
$$\mathbb{E}((X-c) \mathbb{1}_{\{X > c\}}) = \int_{c}^{\infty}(u - c) \mathrm dF(u) = \int_{c}^{\infty} \int_{0}^{u-c} 1 \mathrm dw \mathrm dF(u) = \int_{c}^{\infty} \left(\int_{0}^{u-c} \mathrm dF(u) \right) \mathrm dw$$

but can't make the last connection, especially since in my case integral limit depends on integrand value (using same $u$), hence I'm not even sure that such switching of integral order is possible.

What am I missing?