Corollary 3.7 - Stein & Shakarchi, p85- confusion ![enter image description here]( **Why can we assume $f$ is real-valued?** In fact, I don't see what goes wrong if we assume $f$ is not real valued, as $E_1 \subseteq \mathbb{R}^{d_1}$ still and we may apply previous proposition: _if $E_1, E_2 \subseteq \mathbb{R}^{d_1}, \mathbb{R}^{d_2}$ are measurable respectively, then $E = E_1 \times E_2$ is measurable subset of $\mathbb{R}^{d}$._

It is a common way* to express the proof is based on reductionism.

We ''may'' assume that, because if we know it for real valued function, then extending it to complex valued function is trivial, by noting $f = \Re f + i \Im f$. Note that you _must_ assume $f$ is real valued to define $E_1$ that way. For otherwise, what do we mean by ''$f < a$''?

*Another far too common way: Some fellow students of mine keep saying "WLOG", but this is, obviously, less general.