It is a common exercise to show that if $f\in L^1 $ then$$\int_R |f(x)|d\mu(x) =\int_0^\infty \mu(\{|f|\geq t\})dt $$
In showing this, it is common to make the observation that
$$ \mathbb{1}_{[0,|f(x)|]}(t) = \mathbb{1}_{\{|f|\geq t\}}(x) $$
I can't seem to figure out how these are equal and what the notation means. I am familiar with the indicator function, but not of an indicator function being a function of $ t$ or $x$ here, so perhaps this is part of my misunderstanding.
Why are these two indicator functions equal?
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$\begingroup$Here $\mathbb{1}_P(t)$ is defined to be
$$\mathbb{1}_P(t) = \begin{cases} 1 & P(t) \text{ is true} \\ 0 & P(t) \text{ is false} \end{cases}$$
When $P$ is a set, like $[0, 3]$, then we look at $t \in P$ instead.
So your two functions are
$$ \begin{aligned} \mathbb{1}_{[0, |f(x)|]}(t) &= \begin{cases} 1 & t \in \big [ 0, |f(x)| \big] \\ 0 & \text{else} \end{cases} \\ &= \begin{cases} 1 & 0 \leq t \leq |f(x)| \\ 0 & \text{else} \end{cases} \end{aligned} $$
$$\mathbb{1}_{|f(x)| \geq t}(x) = \begin{cases} 1 & |f(x)| \geq t \\ 0 & \text{else} \end{cases}$$
It should be clear that if we pinky promise to only think about $t \geq 0$, these are actually the same function.
In fact, we can think of this as one function, taking $x$ and $t$ as inputs, call it $\mathsf{blah}(x,t)$. Then our first function is what we get when we fix an $x$ and view this as a function of $t$, and our second function is what we get if we fix a $t$ and view it as a function of $x$. That is:
$$\mathbb{1}_{[0, |f(x)|]}(t) = \mathsf{blah}(x,t) = \mathbb{1}_{|f(x)| \geq t}(x)$$
Now, what happens if we integrate these functions?
Well,
$$\int_0^\infty \mathbb{1}_{[0, |f(x)|]}(t) dt = |f(x)|$$
and
$$\int_0^\infty \mathbb{1}_{|f(x)| \geq t}(x) d \mu (x) = \mu \left ( \{ x \mid |f(x)| \geq t \} \right )$$
(in each case, do you see why?).
Now the integral identity you're asking about is more or less fubini's theorem. We have two variables of interest, and we can integrate in either order!
So we have
$$ \int_0^\infty |f(x)| d \mu(x) = \int_0^\infty \int_0^\infty \mathbb{1}_{[0, |f(x)|]}(t) dt d \mu(x) \overset{\star}{=} \int_0^\infty \int_0^\infty \mathbb{1}_{|f(x)| \geq t}(x) d \mu (x) dt = \int_0^\infty \mu \left ( \{ x \mid |f(x)| \geq t \} \right ) dt $$
where in step $\star$ we used the ideas from earlier in this answer, plus fubini's theorem.
I hope this helps ^_^
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