Can a Nonzero Function Have a Zero Integral Over Any Interval?

[r4nd0m]

Is there a function f: R->R, such that:
$$\forall x \in \mathbb{R}: f(x) \neq 0 \wedge \forall a,b \in \mathbb{R}: \int_a^b f(x) dx = 0$$

I made this problem myself so I don't know, wheather it is easy to see or not. The integral is the Lebesgue integral.
I would say, that there should be such a function, but I don't know how to define it. It has to be discontinuos everywhere, but defining it as something for rational numbers and something else for irrational doesn't help because rational numbers have measure zero.

Any suggestions?

 

[arildno]

Since this would mean that on every interval on R, the function values ought to oscillate about zero, it follows, I think, that f would have to be discontinuous on every point. Otherwise, there would be a single-valued region of finite size about some point, but the integral over that region would be non-zero.

Since therefore f is discontinuous at every point, it is not a Riemann integrable function, and I don't think it is Lebesgue integrable, either.

So, the answer to your question is no.

 

[DeadWolfe]

Why doesn't the characteristic function on the rationals work?

 

[arildno]

Isn't that 0 on reals, and 1 on rationals?
Since f can't be zero anywhere, that doesn't work..

 

[r4nd0m]

A function which is discontinuous everywhere can be Lebesgue integrable - for example the characteristic function of the rationals.

The problem of the characteristic function is that it is equal zero in every irrational point.

 

[arildno]

A function which is discontinuous everywhere can be Lebesgue integrable - for example the characteristic function of the rationals.

Indeed. But that one is zero almost everywhere, which your function is not.
I might be wrong, but I don't think, say, a "Heaviside-like function is integrable (say, 1 on the irrationals, -1 on the rationals)

 

[DeadWolfe]

Ah, I misread the question. I would imagine that if for all a and b:

$$\int_a^b f(x) dx = \int_a^b g(x) dx$$

then f=g a.e.

 

[DeadWolfe]

Indeed. But that one is zero almost everywhere, which your function is not.
I might be wrong, but I don't think, say, a "Heaviside-like function is integrable (say, 1 on the irrationals, -1 on the rationals)

Of course it is, since the rationals and irrationals are both measurable sets.

 

[arildno]

Since I haven't read any measure theory, I'll take your word for that.
EDIT:
Argh, yes of course..

 

[r4nd0m]

nobody knows?

 

[AKG]

Suppose f is not a.e. zero. Then f is either not a.e. non-positive or not a.e. non-negative (or both). Suppose the former, w.l.o.g. Then there is some e > 0, and some bounded measurable set of positive measure, E, such that e < f(x) for x in E. Let $\cap U_n$ be a countable intersection of decreasing bounded open sets containing E with the same measure as E. Then

$$0 = \int _{U_n} f = \int _{U_n - E}f + \int _{E}f > \int _{U_n - E} f + e m(E)$$

The first equality comes from the dominated convergence theorem, together with some facts about bounded open sets, and the assumption that the integral of f is zero on every interval. The next equality is obvious, and so is the following inequality. We can apply the dominated convergence theorem again to choose n so small that

$$\int _{U_n - E} f + e m(E) > 0$$

Doing so gives 0 > 0, a contradiction, so not only must f be sometimes 0, it must be a.e. 0.

 

[r4nd0m]

Suppose f is not a.e. zero. Then f is either not a.e. non-positive or not a.e. non-negative (or both). Suppose the former, w.l.o.g. Then there is some e > 0, and some bounded measurable set of positive measure, E, such that e < f(x) for x in E. Let $\cap U_n$ be a countable intersection of decreasing bounded open sets containing E with the same measure as E. Then

$$0 = \int _{U_n} f = \int _{U_n - E}f + \int _{E}f > \int _{U_n - E} f + e m(E)$$

The first equality comes from the dominated convergence theorem, together with some facts about bounded open sets, and the assumption that the integral of f is zero on every interval. The next equality is obvious, and so is the following inequality. We can apply the dominated convergence theorem again to choose n so small that

$$\int _{U_n \ E} f + e m(E) > 0$$

Doing so gives 0 > 0, a contradiction, so not only must f be sometimes 0, it must be a.e. 0.

I'm sorry, but I couldn't really follow your argument. I think you showed something else - namely that if f>0 (or f<0) almost everywhere then the integral can't be equal zero, which is a well-known fact.

But you skipped the case of f being oscilating (around zero) almost everywhere .

 

[AKG]

I'm sorry, but I couldn't really follow your argument. I think you showed something else - namely that if f>0 (or f<0) almost everywhere then the integral can't be equal zero, which is a well-known fact.

But you skipped the case of f being oscilating (around zero) almost everywhere .

No, I didn't talk about f being positive a.e., I talked about it being not a.e. non-negative. I said:

IF f is not a.e. zero, THEN f is not a.e. non-positive or f is not a.e. non-negative. There are lots of negations in the above, so to see why this is true, look at an equivalent statement, the contrapositive:

IF f is a.e. non-positive and f is a.e. non-negative, THEN f is a.e. zero.

"f is a.e. non-positive" says "{x : f(x) > 0} has zero measure". "f is a.e. non-negative" says "{x : f(x) < 0} has zero measure". If {x : f(x) > 0} has zero measure, and {x : f(x) < 0} has zero measure, then

{x : f(x) > 0} U {x : f(x) < 0}

has zero measure, and this is precisely {x : f(x) != 0}. But "{x : f(x) != 0} has zero measure" is precisely what is meant by "f is a.e. zero". Clear now?

 

[r4nd0m]

Thanks, sometimes I just read to quickly.

The proof looks nice, but since I haven't taken any measure theory yet, I'm not sure about the details, but I'll try to catch that up.

Thanks once again.

 

[AKG]

Admittedly, I left out some details. I feel somewhat guilty for this because my real analysis text does the same and it frustrates me. To make up for it, though, I'll gladly fill in any details you want to see.