$ \mu(|f|\geq \alpha) = \frac{1}{\alpha} ||f||_1$

The conclusion doesn't hold for $f(x) = (1-|x|)_+ = \max(1-|x|,0)$ and if $\mu$ is the Lebesgue measure. The shape of piecewise linear fucntion $f$ is a centred peak, so $||f||_1 = 1$. so that $\mu(|f| \ge a) = 2(1-a) \ne \dfrac1a = \dfrac{||f||_1}{a}$.

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By AM-GM inequality, $$2 \, a(1-a) \le 2 \left(\frac{a+(1-a)}{2}\right)^2 \le 2 \cdot \frac14 = \frac12 < 1,$$ so the last inequality holds.

Let $X:=[0,1]$ and $\mu$ the Lebesgue measure. Let $f:X\to\mathbb{R}$ be as $f(x):=2$ for $x\in[0,1/2]$ and $f(x):=1$ for $x\in(1/2,1]$. Furthermore let $\alpha:=2$

Then $$2\mu([0,1/2]) =1 < 1 +1/4 = \int_0^1 f(x) d\mu(x)$$ which is a counterexample of your equality.

However the proposition is true when you substitute $'='$ by $'\le'$:

$ \mu(\lbrace|f|\geq \alpha\rbrace) \le \frac{1}{\alpha} ||f||_1$.

As others have said this is not necessarily true, but it is true that we are always guaranteed a $\alpha>0$ s.t $\mu\{|f|\geq\alpha\}\leq \frac{\|f\|_1}{\alpha}.$ Regardless of the measure space we know that $\mu\{|f|\geq \|f\|_1\}\leq 1$, so if we choose $\alpha=\|f\|_1$ the inequality will hold.