Measurability of $E=\{(x,y); |f(x)|>y \}$ If $f(x)$ is a Lebesgue integrable function on $R^n$, then is $E=\{(x,y); |f(x)|>y \}\subset R^{n+1}$ measurable?
For each fixed $y$, $E^y=\{x;(x,y)\in E\}$ is measurable and as y increases, $E^y$ decreases, but I don't know whether it will guarantee measurability or not.
 A: The answer to your question is affirmative because of the following facts:


*

*A function $f:T\to \mathbb{R}$ is Lebesgue-integrable (measurable) if and only if its absolute value $|f|$ is also Lebesgue-integrable (measurable). This is a simple exercise based on the decomposition $|f|=f^++f^-$. Second,

*A function $g:T\to \mathbb{R}$ is Lebesgue-measurable if and only if its epigraph $\operatorname{epi} f = \{(x,a): f(x) \leq a\}$ is Lebesgue-measurable (in $T\times \mathbb{R}$). Of course the same can be shown for the hypograph of $f$, that is $\operatorname{hyp} f = \{(x,a): f(x) \geq a\}$. In our case $g=|f|$.

*The graph of a measurable function - that is, the set $\Gamma_f = \{(x,y): y=f(x)\}\subseteq T\times \mathbb{R}$ is a measurable set (in $T\times \mathbb{R}$) and has measure zero. Now notice that the strict hypograph of $f$, that is your set $E$, is $E=\operatorname{hyp} f\setminus \Gamma_f$.


The second statement above can be found in the Book Fixed Point Theory for Decomposable Sets (see by A. Fryszkowski and it comes from a result by Castaing and Valadier [1] where the authors state interesting relevant results for multifunctions (mutli-valued measurable functions).
I would also recommend the book of Rockafellar and Wets [2] where in Chapter 14 the authors elaborate on measurability of multifunctions and to a great extent they use the original results of Castaing such as Castaing representations.
[1] C. Castaing and M. Valadier, Convex analysis and measurable multifunctions, Lect. Notes in Math. 580, Springer Verlag, Berlin, 1977.
[2] R.T. Rockafellar, R.J.-B. Wets, Variational analysis, Springer, 3rd edition, 2009.
A: Since you know that if $E $ is measurable, then so is $E \times \Bbb {R} $, you can do the following:
Define
$$
F : \Bbb {R}\times \Bbb {R} \to \Bbb {R}, (x,y)\mapsto f (x).
$$
Then, for every $M\subset \Bbb {R} $, we have $F^{-1}(M)=f^{-1}(M )\times \Bbb {R} $, from which it easily follows that $F $ is measurable and hence also $|F|$.
But the map $G: \Bbb {R}^n \times \Bbb {R}, (x,y)\mapsto y $ is continuous and hence also measurable. Hence so is $|F|-G $.
But we finally have
$$\{(x,y)\mid |f (x)|>y\} = \{(x,y)\mid (|F|-G)(x,y)>0\} = (|F|-G)^{-1}((0,\infty)) , $$
where the set on the right is measurable.
This argument generalises to many different settings,  in contrast to the "epigraph" approach. For example if you see interested in the set where $|f (x)|>\sin (y) $.
