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?

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    $\begingroup$ I'd say the Radon-Nikodým theorem is pertinent. $\endgroup$ Jul 15, 2015 at 10:08
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    $\begingroup$ @DanielFischer I don't see the connection.. I'd love it if you could expand on your comment! :-) $\endgroup$
    – Ant
    Jul 15, 2015 at 10:15
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    $\begingroup$ 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. $\endgroup$ Jul 15, 2015 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, 2015 at 10:20
  • $\begingroup$ It's also can be deduced from Lebesgue differentiation theorem $\endgroup$
    – Mher
    Jul 15, 2015 at 11:25

3 Answers 3


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.

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    $\begingroup$ +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, 2015 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, 2015 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. $\endgroup$ Jul 15, 2015 at 10:41
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    $\begingroup$ The question is "what's the name of the theorem". As correct as this answer is, it doesn't give the name. $\endgroup$ Jul 15, 2015 at 18:02
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    $\begingroup$ @BrianVandenberg Well if the theorem does not have a name, I can't do much about it, can I? :-) $\endgroup$
    – Ant
    Jul 15, 2015 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, 2015 at 10:10
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    $\begingroup$ 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, 2015 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).


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