Scorza-Dragoni theorem (at least the version I have used) says that if you have a function $f : \Omega \times \mathbb{R}^{N} \longrightarrow \overline{\mathbb{R}}$ which satisfies:

i) $x \rightarrow f(x,v)$ is measurable for all $v \in \mathbb{R}^{N}$

ii) $v \rightarrow f(x,v)$ is continuous for almost every $x \in \Omega$

iii) $f$ is locally integrable

Then for any positive $\epsilon$ there exists a compact subset of $\Omega,$ say $K,$ such that $f$ restricted to $K \times \mathbb{R}^{N}$ is a continuous function and $\vert K^{c} \vert < \epsilon$

I tried to prove this before looking for it somewhere and came up with a stupidly simple proof, which confuses me, because if this were right, the theorem wouldn't have its own name :S

My reasoning is: As we have a measurable function, we can approximate it pointwise by step functionts which, in turn, can be approximated pointwise by continuous functions almost everywhere. Theorefore, by Egorov's theorem we can find $K$ such that $\vert K^{c} \vert < \epsilon$ and such that those continuous functions converge uniformly. The limit on $K$ is then continuous and clearly coincides with the restriction of $f.$

There must be some subtle technical point I am omiting, because I guess the proof of a named theorem cannot be this silly. Where is my mistake?

Thanks for your help

EDIT: Added the almost everywhere that was missing

  • $\begingroup$ What's the nature of $\Omega$, and which measure is at play in "for almost every $x\in\Omega$"? $\endgroup$ Commented Oct 17, 2015 at 15:54
  • $\begingroup$ "we have a measurable function, we can approximate it pointwise by step functionts which, in turn, can be approximated pointwise by continuous functions": You need to carefully state and prove this fact. It's not true pointwise, though it is true pointwise almost everywhere. $\endgroup$ Commented Oct 17, 2015 at 15:56
  • $\begingroup$ certanly, that's true. Stil, the argument applies in the same way $\endgroup$
    – Qwertuy
    Commented Oct 17, 2015 at 16:21

1 Answer 1


You are applying Egoroff's theorem to the function $x \mapsto f(x,y)$. This means that you get a compact set $K_y$ for which $f(\cdot,y) \colon K_y \to \mathbb{R}$ is continuous. Since the compact set depends on $y$, you need to show that you can pick one that works for a.e. $y$; it is not obvious to me how to do it (we have more than countably many $y$'s), and I think this is what your proof fails to accomplish. Let me know what you think!

  • 1
    $\begingroup$ What if one tries to apply the Egorov/Lusin machine to the map $x\mapsto f(x,\cdot)$ from $\Omega$ to $C(\Bbb R)$? There is no loss in assuming $f$ to be bounded, so one could take the range space to be $C_b(\Bbb R)$ equipped witth the uniform norm. One would have to start by re-defining $f(x,\cdot)$ for a null set of $x$ values to ensure that $f(x,\cdot)\in C_b(\Bbb R)$ for all $x\in\Omega$. $\endgroup$ Commented Oct 17, 2015 at 17:30
  • $\begingroup$ Mh, that's an interesting point of view! I have to think about it.. The most general Lusin's theorem I know requires the target space to be second countable I think.. Is this true for $C_b(\mathbb{R})$? I guess it is reason why you want the function to be bounded, but is that enough? The way I would prove second countability is by Stone-Weierstrass, but if I remember the statement correctly for locally compact hausdorff spaces you can only approximate uniformly functions in $C_0$. Am I wrong? $\endgroup$
    – Giovanni
    Commented Oct 17, 2015 at 17:45
  • $\begingroup$ You're right, $C_0(\Bbb R)$ is more appropriate from the second-countability point of view. $\endgroup$ Commented Oct 18, 2015 at 0:52

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