It happens very often in physics that we find relations like:

$$\int_V f(x) dx = \int_V g(x) dx$$

for an arbitrary volume $V$. From this we usually say "Since the volume is arbitrary, the integrands have to be equal" and we conclude $f(x) = g(x) $.

Funnily enough, this is often stated without even mentioning the name of the theorem that allows us to do so.

Intuitively, it's clear why is it so; but what's the theorem that formally allows us to do that?

  • 5
    $\begingroup$ I'd say the Radon-Nikodým theorem is pertinent. $\endgroup$ – Daniel Fischer Jul 15 '15 at 10:08
  • 1
    $\begingroup$ @DanielFischer I don't see the connection.. I'd love it if you could expand on your comment! :-) $\endgroup$ – Ant Jul 15 '15 at 10:15
  • 3
    $\begingroup$ If you define $\nu(V) = \int_V f(x)\,dx$ (for measurable $V$), you have an absolutely continuous (wrt Lebesgue measure) [complex/signed] measure $\nu$, $f$ is its Radon-Nikodým derivative. Do the same for $g$, you get the same measure by assumption, hence the same Radon-Nikodým derivative, hence $f = g$ almost everywhere. $\endgroup$ – Daniel Fischer Jul 15 '15 at 10:20
  • $\begingroup$ @Ant The Radon-Nikodym theorem states that under certain conditions there is a somehow "unique" derivative. This uniqueness (a.e. w.r.t.) ensures that $f=g$ a.e. w.r.t. $\endgroup$ – drhab Jul 15 '15 at 10:20
  • $\begingroup$ It's also can be deduced from Lebesgue differentiation theorem $\endgroup$ – Mher Jul 15 '15 at 11:25

Let $(X,\mathcal{E},\mu)$ be a measure space. In general, if $f,g:X\to\mathbb{R}$ are both measurable with respect to $\mathcal{E}$, integrable, and satisfy $$ \int_A f\,\mathrm d\mu =\int_A g\,\mathrm d\mu,\qquad \text{for all } A\in\mathcal{E}, $$ then $f=g$ almost everywhere with respect to $\mu$. In fact, it is enough to require the identity for all $A$ in some $\cap$-stable generator of $\mathcal{E}$.

In particular, if you let $(X,\mathcal{E},\mu)=(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ where $\lambda$ is the Lebesgue measure, then if $f,g:\mathbb{R}\to\mathbb{R}$ are Borel measurable, integrable, and satisfy $$ \int_A f\,\mathrm d\lambda=\int_Ag\,\mathrm d\lambda,\qquad\text{for all }A\in\mathcal{B}(\mathbb{R}), $$ you can conclude that $f=g$ almost-everywhere with respect to $\lambda$. Again, it is enough to require the identity e.g. for all $A$ of the form $[a,b]$, $a<b$. Note that you can never be sure of pointwise identity.

  • 2
    $\begingroup$ +1 thank you, I love this. Two questions though: 1) If we require $f$ and $g$ to be continuous, do we have pointwise identity? (sidenote: is it possible to define continuity in a general measure space or we need also a topological structure? ) and 2) Does this theorem have a name or is a minor result without name? :-) Thank you! $\endgroup$ – Ant Jul 15 '15 at 10:13
  • $\begingroup$ If you require continuity, then they are equal, as their difference is continuous and zero almost-everywhere. $\endgroup$ – Omry Jul 15 '15 at 10:16
  • $\begingroup$ @Ant: You need some kind of topological structure in order to talk about open sets and continuity. I have never heard this theorem go under any name by itself, but it's a consequence of the uniqueness of densities which again is a consequence of the uniqueness theorem of measures. $\endgroup$ – Stefan Hansen Jul 15 '15 at 10:41
  • 4
    $\begingroup$ The question is "what's the name of the theorem". As correct as this answer is, it doesn't give the name. $\endgroup$ – Brian Vandenberg Jul 15 '15 at 18:02
  • 3
    $\begingroup$ @BrianVandenberg Well if the theorem does not have a name, I can't do much about it, can I? :-) $\endgroup$ – Ant Jul 15 '15 at 18:11

This statement actually isn't true in general. Consider the following functions $f,g$ where $f$ is the zero function and we define $g(x) = 0$ for $x \neq 0$ and $g(0)=1.$ Clearly the two functions are not the same, however for any interval $[a,b],$

$$ \int_a^b f(x) \ dx = \int_a^b g(x) dx = 0. $$

Using a similar argument you can construct a counterexample in $\Bbb R^3.$

This statement is true however, if we assume that $f$ and $g$ are continuous. This is because if $f(x) \neq g(x),$ for some $x \in \Bbb R^3,$ then there is an open neighbourhood for which $f$ and $g$ are not equal (and by continuity one is always larger than the other). Integrating over this set gives a contradiction.

I don't know if this result has a name though and I wouldn't be surprised if it didn't. Important theorems are usually given a name as they are common referenced, but smaller results like these usually aren't given one.

Edit: Some conditions for $V$ may also be necessary. A sufficient condition (which my proof sketch uses) if $V$ to be open.

  • $\begingroup$ +1 Thank you, this is nice. I wonder if without the continuity assumption we can conclude $f=g$ outside of a set of lebesgue measure zero? $\endgroup$ – Ant Jul 15 '15 at 10:10
  • 1
    $\begingroup$ I would strongly advocate the view that frequently "value at specific point" is meaningless at points of discontinuity. In $L^p$ spaces that is not even defined, and likely so for distributions. There are also physical reasons. $\endgroup$ – Blazej Jul 15 '15 at 10:47

I think the type of theorem you would formally need to use depends on type of functions.

If they are continuous you would just prove it by contradiction: assume that there is a point where they are not equal. By continuity there is whole neighbourhood of that point where $f-g$ has constant sign. Integrating $f-g$ on that volume leads to absurd.

If functions you are considering are in $L^p$, $p \in [0, \infty)$ then linear span of characteristic functions of finite volumes are dense in the dual space $L^q$ so $f-g$ must be a zero vector. Since those are just equivalence classes of functions equal a.e. this doesn't contradict example given by ctoi. Note however that when in practice (in physics) we use discontinuous functions, values at any particular points are irrelevant - all action is in open sets (e.g. in QM wave function in specific point is meaningless, it's the integral that matters).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.