Let f(x, y) be a real valued function of the real variables x and y.
For any x, f(x, .) is measurable. For any y, f(., y) is measurable.

Is f(x, y) measurable? Proof? Counter-example? Pointers to resources?

If the answer is no, would it help if f(., y) is also continuous?

Thanks.

Check "Counterexamples in Analysis" by B. Gelbaum and J. Olmsted; ch.
10, example 21. There presented an example (due to Sierpinski) of an
unmeasurable set C in R^2 such that its intersection with any straight
line contains no more than 2 points.

Let f be the characteristic function of C. Then f(x,.) equals 0 for
every y in R except for no more then 2 points (where it equals 1). The
same is true for f(.,y). So f(x,.) and f(.,y) are measurable. But
f(.,.) is unmeasurable (since C is unmeasurable).

"...would it help if f(., y) is also continuous?" I dont know.

Here is a proof that if f(x,y) is continuous in the first coordinate
and measurable in the second one, then f is measurable in both
coordinates. Fix a < b. Let S= all (x,y) such that f(x,y) is between a
and b. We wish to prove that S is measurable. For rationals p < q ,
let Spq = all (x,y) such that for all rationals r between p and q 
f(r,y) is between a and b and x is between p and q. Then Spq is the
countable intersection of measurable sets since f is measurable in the
second coordinate. But S is the countable union of all Spq for p < q
since f is continuous in the first coordinate. So S is measurable.

Please, could You explain why S is a union of Spq. Am i right that 
S = { (x,y) | a < f(x,y) < b } and
Spq = { (x,y) | p < x < q and a < f(r,y) < b for all rational r such
that p < r < q }?
I cannot understand why Spq lies in S. (Of course, it is easy to see
that  a <= f(x,y) <= b when (x,y) is in Spq.)

To lambertch: You have a point. Here's a way around the problem: Yes,
S= all (x,y) such that a< f(x,y) < b. For rationals p < q and positive
integer n define Spqn to be all(x,y) such that x is in [p,q] and for
all rationals r in [p,q] a+1/n<= f(x,y) <= b - 1/n. Then if (x,y) is
in S it is in some Spqn. And if (x,y) lies in some Spqn then a+1/n <=
f(x,y) <= b-1/n by the continuity of f in the first coordinate. So
f(x,y) is in S. So S is the countable union of measurable sets.

Correction: .... for all rationals r in [p,q] a + 1/n <= f(r,y) <= b - 1/n .

lophar, berkeleychocolate:

Thank you both for your answers. 

I had the book "Counterexamples in Analysis" but I have yet to locate
it. It's good to know that counter example.

I have a question about berkeleychocolate's proof. Why is Spqn
measurable? It is  still a set of real numbers, not rational numbers,
right?  To make things clearer, let's forget about y and consider f(x)
only. Then your proof would constitute a proof for "continuous
functions are measurable," wouldn't it? Let f(x) = x; it seems to me
that you are getting a countable union of countable points for an open
interval -- I don't know what I'm missing.

Best,

lambertch

Certainly every continuous function is measurable since every interval
is measurable. The above proof relies basically on the fact that if a
function is continuous then an inequality is valid for all x if it is
valid for all rationals r. No, intervals are not countable.

Let me try to be more detailed in the above proof. Let M(A,B,r) = {y |
A <= f(r,y) <= B}. Then M(A,B,r) is measurable for all A,B,r since f
is measurable in the second coordinate, y. Let N(A,B,p,q) = the
intersection over all rationals r in the closed interval [p,q] of
M(A,B,r). Then N(A,B,p,q) is measurable since the countable
intersection of measurable sets is measurable. Note that N(A,B,p,q) =
{ y | for all rationals r in [p,q] A<=f(r,y)<=B}. Let R(A,B,p,q) = the
cross product of the sets [p,q] and N(A,B,p,q). Then R(A,B,p,q) is
measurable in the plane since it is the cross product of two
measurable sets. Note that R(A,B,p,q) = {(x,y) |  p<=x<=q and for all
rationals r if p<=r<=q then A<=f(r,y)<=B}. Finally let S(a,b) be the
union over all rational nbrs p and q and positive integers n of
R(a+1/n,b-1/n,p,q). Since each R(a+1/n,b-1/n,q,q) is measurable and we
are taking a countable union to get S(a,b), S(a,b) is also measurable
in the plane.

Let S = {(x,y) | a<f(x,y)<b}. All we have to show now is that S(a,b)
is actually S. First suppose (x,y) is in S. Then clearly there is some
positive integer n such that a+1/n<=f(x,y)<=b-1/n. Now by the
continuity of f in the first coordinate there is some d>0 such for all
x' in (x-d,x+d) a+1/n<=f(x',y)<=b- 1/n. Pick rationals p and q such
that x-d<=p<x<q<=x+d. Then for all x'in [p,q] and therefore for all
rationals x' in [p,q] a+1/n<=f(x',y)<=b-1/n. This says that (x,y) lies
in R(a+1/n,b-1/n,p,q) and therefore in S(a,b).

The other inclusion is easier. If (x,y) lies in some
R(a+1/n,b-1/n,p,q), pick a sequence of rationals rm converging to x.
Then since for all m a+1/n<=f(rm,y)<=b-1/n, it follows by continuity
of f in the first coordinate that a+1/n<=f(x,y)<=b-1/n and so
a<f(x,y)<b.

If there is an error in this proof, I don't see it.

berkeleychocolate-ga: Thank you so much for your detailed proof! Now I get it!

For whatever its worth: I've written a letter of recommendation to Google Answers. 

Best,

lambertch-ga

Hi berkeleychocolate-ga: I'm writing a technical paper and the result
you gave constitutes a lemma with a complete proof. I would like to
acknowledge your contribution in the paper. How should I do this?
Please feel free to email me at lambertch@hotmail.com.

Thanks,

lambertch-ga