Do this integral really means an integral? (random set representation of a fuzzy set) If yes, can someone explain to me intuitivily why this is true?

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Let's suppose that I have a crisp set like:

\begin{equation} A^\alpha = \{x \mid A(x) \geq \alpha \} \end{equation}

and let's define a function

\begin{equation} f_\alpha(x) = \begin{cases} 1, & \text{se}\ x \in A^\alpha \\ 0, & \text{se}\ x \notin A^\alpha \end{cases} \end{equation}

Can I write the value of A(x) as this?

\begin{equation} A(x) = \int_0^1 f_\alpha(x)d\alpha \end{equation}

I saw this in a book and I can't make sense of it. I read in the Klin and Yuan book that this integrated representation doesn't mean literally a integration, but on Fuzzy Logic and Mathematics: A Historical Perspective from Bělohlávek the equation appears like a normal integration. I can't form a picture in my mind of why this is true, and how it is possible of function that returns only 1 and 0 can have a integral that represents a fuzzy set.

I am posting a printscreen of the passage if the question isn't entirely clear.

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Maybe a key thing to remember is that you're integrating with respect to alpha, not with respect to x. I've included a diagram that hopefully helps you get a visualisation of what's happening.

The alpha values can be thought of as horizontal cuts. Consider a value $x_1$ so that $A(x_1) = \alpha$. For all values $x$ below $x_1$, $F_\alpha (x) = 1$, then for all values $x$ above, $F_\alpha(x) = 0$, so effectively you'll be summing up the $(1 \cdot d\alpha)$ until you hit $x$, which will just return the value of $\alpha = A(x_1)$.

Graphs of $A(x)$ and corresponding $f_\alpha$: