20
$\begingroup$

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?

$\endgroup$
0

6 Answers 6

13
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ $[a,b]\setminus K$ is not necessarily open. $\endgroup$
    – nullUser
    Oct 4, 2012 at 1:47
13
$\begingroup$

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 $dm$.

$\endgroup$
8
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
7
$\begingroup$

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 $n+I,\;n\in \mathbb{Z}$.

$\endgroup$
2
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

then

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

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

Proof:

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

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

Thus

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

$\displaystyle \circ$

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 $\displaystyle 0$.

Thus

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

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 $\displaystyle E$, $\displaystyle \int_{E} f = 0$.

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
  • 4
    $\begingroup$ 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
0
$\begingroup$

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).

$\endgroup$

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.