Krdytkyu

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?

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,
$$ int_O {f(t) dt} = sumlimits_{n = 1}^infty {int_{c_n }^{d_n } {f(t) dt} } = 0 $$.

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.

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 $A=Bcup Z$ 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 $dm$.

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

Let D be all the measurable sets $Usubseteq I=[0,1]$ such that $intop_U f(t) = 0$ (the function $fmid_I$ is in $L_2$ so it is also in $L_1$, so I assume this from now). $Iin D$ and if $Asubseteq Bsubseteq I$ 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 $Psubseteq D$). 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 $Psubseteq D$ then $sigma(P)subseteq D$. The sigma algebra generated by P is the Borel algebra.

Now look on the set $A={xin I mid f(x)geq 0}$. 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 $n+I,;nin mathbb{Z}$.

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

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.