I have been thinking on and off about a problem for some time now. It is inspired by an exam problem which I solved but I wanted to find an alternative solution. The object was to prove that some sequence of functions converges weakly to zero in $L^2$.

I managed to show (with some help) that the limit $f$ (of a subsequence) satisfies $\int_0^x f \ dm=0$ for all $x>0 $. From this I want to conclude that $f=0$ a.e. I can do this with the fundamental theorem of calculus in its Lebesgue version but there ought to be a more elementary proof.

Can someone here help me out?


6 Answers 6


Indeed, as you expected, a simple proof of the result can be found; see Theorem 2.1 in this useful note on absolutely continuous functions.

EDIT: Since this is a quite important result, it is worth giving here the proof in detail. The proof below is essentially the one given in the link above, but somewhat shorter.

Theorem. If $f$ is integrable on $[a, b]$ and $\int_a^x {f(t) dt} = 0$ $\forall x \in [a,b]$, then $f = 0$ a.e. on $[a, b]$.

Proof. An open subset $O$ of $[a,b]$ is a countable union of disjoint open intervals $(c_n, d_n)$; hence, $$ \int_O {f(t) dt} = \sum\limits_{n = 1}^\infty {\int_{c_n }^{d_n } {f(t) dt} } = 0 $$.
If $K$ is a closed subset of $[a,b]$, then $$ \int_K {f(t) dt} = \int_a^b f(t)dt - \int_{(a, b) \setminus K} f(t)dt = 0 - 0 = 0, $$

since $ (a, b) \setminus K $ is open.

Next let $E_ + = \{ x \in [a,b]:f(x) > 0\}$ and $E_ - = \{ x \in [a,b]:f(x) < 0\}$. If $\lambda(E_+) > 0$, then there exists some closed set $K \subset E_+$ such that $\lambda(K) > 0$. But $\int_K {f(t){\rm d}t} = 0$, hence $f=0$ a.e. on $K$. This contradiction shows that $\lambda(E_+) = 0$. Similarly, $\lambda(E_-) = 0$. The theorem is thus established.

  • 2
    $\begingroup$ $[a,b]\setminus K$ is not necessarily open. $\endgroup$
    – nullUser
    Oct 4, 2012 at 1:47

It is sufficient to prove that $f$ is zero almost everywhere on any bounded interval.

(1) By additivity it is easy to see that $$\int_a^bf(x)dx=\int_0^bf(x)dx - \int_0^af(x)dx$$ for all bounded intervals $(a,b)$ (and also for $[a,b)$, $(a,b]$ and $[a,b]$).

(2) Using (1) it is easy to see that $$\int_Bf(x)dx=0$$ for any bounded Borel measurable set.

(3) Any Lebesgue measurable set $A$ is of the form $A=B\cup Z$ where $B$ is a Borel measurable set and $Z$ is a set of measure zero. Hence, by (2) we acheive $$\int_A f(x)dx= 0$$ for any bounded Lebesgue measurable set $A$.

(4) Now look at the sets $A_+(n)=\{x:f(x)>0\}\cap[-n,n]$ and $A_-(n)=[-n,n]\setminus A_+(n)$. Assuming $f$ is measurable these sets are also measurable and by (3) $$\int_{A_\pm(n)}f(x)dx=0$$ EDIT: and hence $f=0$ almost everywhere.

Please forgive me if I write $dx$ for the Lebesgue measure which I presume is what you refer to as $dm$.

  • $\begingroup$ Yes, that is what I mean by $dm$. Very nice argument. $\endgroup$
    – Johan
    Jan 3, 2011 at 23:00
  • $\begingroup$ How elementary is a proof of (3)? $\endgroup$
    – Aryabhata
    Jan 4, 2011 at 22:31
  • $\begingroup$ @Moron: Well, perhaps as elementary as the existence of the Lebesgue measure. $\endgroup$ Jan 5, 2011 at 6:48
  • $\begingroup$ Never mind, I found a proof. It seems to be actually quite similar to my answer. $\endgroup$
    – Aryabhata
    Jan 5, 2011 at 7:16
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    $\begingroup$ @Moron: (2) is easier than that by (1) $\int_I f dx=0$ for all bounded intervals $I$ and those intervals ($\sigma-$)generates the bounded Borel sets. $\endgroup$ Jan 9, 2011 at 7:44

I think you can use Dynkin's lemma (if you call this "more elementary").

Let D be all the measurable sets $U\subseteq I=[0,1]$ such that $\intop_U f(t) = 0$ (the function $f\mid_I$ is in $L_2$ so it is also in $L_1$, so I assume this from now). $I\in D$ and if $A\subseteq B\subseteq I$ are in $D$ then $B-A \in D$. If $A_i \subseteq I$ is an increasing sequence in D then $\bigcup A_i \subseteq I$ is also in D (by the DCT). This shows that D is a Dynkin system.

Let P be all the open intervals in I (so $P\subseteq D$). P is not empty and an intersection of two open intervals are open, so P is closed under finite intersection, hence it is a pi system.

Dynkin's lemma says that if P is a pi system and D a dynkin system such that $P\subseteq D$ then $\sigma(P)\subseteq D$. The sigma algebra generated by P is the Borel algebra.

Now look on the set $A=\{x\in I \mid f(x)\geq 0\}$. This is a Lebesgue measurable set, so up to a zero measure set it is Borel measurable set $A'$. Since $\intop_{A'} f(t) = 0$ and f is non negative there, then f is zero almost every where in A'. The same argument work for when f<0, so you get that f is zero almost everywhere in $I$. Now do this for all of $n+I,\;n\in \mathbb{Z}$.

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    $\begingroup$ Very nice. The last step is unnecessary if $f$ is Borel measurable, and even if it is only Lebesgue measurable we could use the fact that a Lebesgue measurable function $f$ is a.e. equal to a Borel measurable function $\tilde{f}$, and consider $\tilde{f}$ instead. $\endgroup$ Jan 3, 2011 at 17:48
  • $\begingroup$ It was not obvious to me that the integral over a Borel set must be zero, I added a proof of that to my answer. $\endgroup$
    – Aryabhata
    Jan 8, 2011 at 23:10

I believe here is an elementary proof (if you are willing to call dominated convergence theorem as elementary).

First a lemma:

Lemma: let $\displaystyle A$ be a bounded measurable set and let $\displaystyle f \in L(A)$. If $A_n \subset A$ is a sequence of measurable sets such that

$$ A_1 \supset A_2 \supset A_3 \supset \dots$$

and $$\lim_{n \to \infty} m(A_n) = 0$$


$$\lim_{n \to \infty} \int_{A_n} f \ \text{dm} = 0$$

($\displaystyle m(T)$ is the lebesgue measure of $\displaystyle T$).


It is well known (and has an elementary proof) that $\displaystyle X = \bigcap_{n=1}^{\infty} A_n$ is measurable and $\displaystyle m(X) = \lim_{n \to \infty} m(A_n) = 0$.

Now define a sequence of (summable) functions

$\displaystyle f_n(x) = \begin{cases} 2 f(x) & x \in A_n \\ f(x) & \text{otherwise} \end{cases}$

Now $\displaystyle |f_n(x)| \le |2f(x)|$ and $f_n \to f$ almost everywhere.

The set of points $\displaystyle S$ where $f_n(x) \to f(x)$ is not true, satisfies $\displaystyle S \subset X$ and hence is measurable and $\displaystyle m(S) = 0$.

By the dominated convergence theorem we have that

$$\lim_{n \to \infty} \int_{A} f_n = \int_{A} f$$

But we have that

$$\int_{A} f_n = \int_{A} f + \int_{A_n} f$$


$$\lim_{n \to \infty} \int_{A_n} f = 0$$

$\displaystyle \circ$

Note that if $\displaystyle f$ was bounded, then there is a much simpler proof of the above lemma, which does not make use of the dominated convergence theorem.

Now back to the original problem.

Let $\displaystyle P_n = \{ x : f(x) \ge \frac{1}{n} \}$.

If the set $\displaystyle P = \{x : f(x) \gt 0\} = \bigcup P_n$ is of positive measure, then there is an $\displaystyle n$ for which $\displaystyle m(P_n) \gt 0$. Now if $\displaystyle P_n$ is unbounded, there is some $\displaystyle M$ for which $\displaystyle m(P_n \cap [M, M+1]) \gt 0$. Call that set $\displaystyle A$.

Notice that $\displaystyle \int_{A} f \ge \frac{m(A)}{n} \gt 0$.

Now give an integer $\displaystyle k \gt 0$, there is an open set $\displaystyle G_k \supset A$ such that $\displaystyle m(G_k-A) \lt \frac{1}{k}$.

Note that we can choose the $\displaystyle G_i$ such that $\displaystyle G_1 \supset G_2 \supset G_3 \supset \dots$, by taking $\displaystyle G'_k = \bigcap_{i = 1}^{k} G_i$.

Now the sequence of sets $\displaystyle A_k = G'_k -A$ satisfies the conditions of the above lemma,

we also have

$$\int_{G'_k} f = \int_{A} f + \int_{A_k} f$$

Now since $\displaystyle G'_{k}$ is a countable union of intervals, we have that $\displaystyle \int_{G'_k} f = 0$, since over every interval, the integral of $\displaystyle f$ is $\displaystyle 0$.


$$\int_{A} f + \int_{A_k} f = 0$$

Taking limits, and applying above lemma, we get

$$\int_{A} f = 0$$

A contradiction. Similarly, we can show that negative set of $\displaystyle f$ is of measure $\displaystyle 0$ (or just consider $\displaystyle -f$).

Hence $\displaystyle f = 0 \ \text{a.e}$

Note: Since this answer almost proves two claims made by other answers, I am including a sketch of proof of those here:

Claim 1) For any measurable set $\displaystyle A$, there is a Borel Set $\displaystyle B \supset A$ such that $\displaystyle m(B) = m(A)$.

For a proof of that, consider the $\displaystyle G'_{k}$ above. $\displaystyle B = \bigcap_{k=1}^{\infty} G'_{k}$ is a Borel set such that $\displaystyle m(B) = m(A)$, as $\displaystyle m(B) = \lim_{k \to \infty} m(G'_{k}) = m(A)$.

Claim 2) For the $\displaystyle f$ in the problem, for any Borel set $\displaystyle B$, $\displaystyle \int_{B} f = 0$.

The proof above actually shows that for any measurable set $\displaystyle E$, $\displaystyle \int_{E} f = 0$.

  • $\begingroup$ Nice!! But you forgot to mention (I think) how you make $P$ out of $P_n$. $\endgroup$
    – JT_NL
    Jan 4, 2011 at 19:21
  • $\begingroup$ @Jonas: Thanks! edited. $\endgroup$
    – Aryabhata
    Jan 4, 2011 at 19:58
  • $\begingroup$ Nice argument. But I do not understand your use of Lusin's theorem in the alternative solution. How do you conclude that $f$ must be continuous at some point? That the positive set has positive measure is not enough for general functions of course but I do not see how the fact that the function integrates to zero over intervals helps. $\endgroup$
    – Johan
    Jan 5, 2011 at 9:54
  • $\begingroup$ @Johan: I was mistaken, I missed the 'when restricted to" part of the theorem. In fact the characteristic function of rationals is a counterexample to what I had claimed. I have deleted that portion from the answer. $\endgroup$
    – Aryabhata
    Jan 5, 2011 at 19:18

If $F(x)=\int_0^x f(t) dt=0$ everywhere, then $F'(x)=0$ for all $x$. Since $f$ is locally integrable, $F'(x)=f(x)$ almost everywhere. Hence $f(x)=0$ almost everywhere.

  • $\begingroup$ This uses the fundamental theorem of calculus, which the OP said he wanted to avoid. $\endgroup$ Jan 3, 2011 at 16:33
  • $\begingroup$ You are right; I did not read closely. Since the OP is working with $L^2$, Lebesgue integration is unavoidable; I don't see the point of trying to avoid this basic theorem in the theoy. $\endgroup$
    – TCL
    Jan 3, 2011 at 16:39
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    $\begingroup$ The only point is that the fundamental theorem of calculus for Lebesgue integrals is treated quite late in the expositions I have seen, after one has proven a great deal of other properties of the integral. But the statement I am seeking to prove seems so simple and intuitive that there ought to be a simple proof. $\endgroup$
    – Johan
    Jan 3, 2011 at 22:55

Here is yet another argument. The only (somewhat) non-elementary fact that I use is that for each $f \in L^1 ([0,A])$ and $\varepsilon > 0$, one can find a continuous, compactly supported function $g \in C_c(\Bbb{R})$ with $\mathrm{supp} \, g \subset [0,A]$ such that $\| f - g\|_{L^1([0,A])} < \varepsilon$.

Without loss of generality, I can assume that $f$ is real-valued (otherwise, apply the argument below to the real and imaginary part separately).

As others have already noted, we have $\int_a^b f(x) \, d x = \int_0^b f(x) \, d x - \int_0^a f(x) \, dx = 0$ for any $a \leq b$. This implies that $\int_\Omega f(x) \, d x = 0$ for any set $\Omega \subset \Bbb{R}$ which is a countable, disjoint union of intervals.

It is not hard to see that this holds for every open set $\Omega \subset \Bbb{R}$. In case you don't want to read through the linked question, I provide a short proof below.

It is enough to prove that $f = 0$ almost everywhere on $[0,A]$, for fixed but arbitrary $A > 0$. Suppose this is not so; then $\varepsilon := \|f\|_{L^1([0,A])} > 0$, so that by the property mentioned above there is a function $g \in C_c (\Bbb{R})$ with $\mathrm{supp} \, g \subset [0,A]$ such that $\| f - g\|_{L^1([0,A])} < \varepsilon / 4$.

Note that $\Omega := \{ x : g(x) > 0 \}$ is open, and $\Omega \subset [0,A]$. Hence, $$ \bigg| \int_\Omega g(x) \, dx \bigg| = \bigg| \int_\Omega g(x) - f(x) \, d x \bigg| \leq \int_0^A |g(x) - f(x)| \, d x \leq \frac{\varepsilon}{4}. $$ In exactly the same way, we also get $|\int_{\Omega'} g(x) \, d x| \leq \varepsilon / 4$ for $\Omega' := \{x \colon g(x) < 0\}$.

Therefore, $\|g\|_{L^1([0,A])} \leq \varepsilon / 2$, so that $\varepsilon = \|f\|_{L^1([0,A])} \leq \|f - g\|_{L^1} + \|g\|_{L^1} \leq \frac{3}{4} \varepsilon$, which is the desired contradiction.

Proof that each open set is the countable disjoint union of open intervals: Each set $\Omega \subset \Bbb{R}$ is the disjoint union of its connected components, each of which is (in the case of an open set $\Omega$) an open interval; furthermore, since $\Bbb{R}$ is separable, any collection of disjoint, non-empty open sets has to be countable. (Slightly different argument: Each of the connected components $I \subset \Omega$ contains a rational number $q_I \in I$, and $q_I \neq q_J$ for $I \neq J$, since the components are disjoint; hence, the map $I \mapsto q_I$ is injective from the set of connected components of $\Omega$ into the rational numbers).


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