Boundness of Hilbert Transform - Finding an absolute constant for $0<p<1$

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I am doing a course in Harmonic analysis and we are looking into the Hilbert Transform. Boundness properties for $L^p$ when $p \in (1, +\infty)$ is not unfamilar. We have gone through the classical interpolation theorems (Marcinkiewicz etc) and touched upon Calderon Zygmund operators and distributions.

I have now come across a question to show that there exists and absolute constant $C_1 > 0$ such that

$ (\int_I|H(f)(x)|^pdx)^{1/p} \leq \frac{C_1}{(1-p)^{1/p}}|I|^{(1-p)/p} \| f\|_{L^1(\mathbb{R})} $ for all $f \in L^1(\mathbb{R}) $ and $p \in (0,1)$

where $H(f)(x)$ is the Hilbert Transform and $I \subset \mathbb{R}$ a bounded interval.

The hint is to note that there exists and absolute constant $C_1 > 0$ such that for all $f \in L^1(\mathbb{R})$ one has

$$|\{ x \in I : |H(f)(x)|> \alpha\}| \leq \min\{|I|, \alpha^{-1}C_1\| f\|_{L^1(\mathbb{R})} \}$$

This $p \in (0,1)$ is really throwing me off and I do not know where to start. Any suggestions on the approach to take or is it even elementary? I feel like I am missing something obvious here.

Thank you very much.