Show that f(t) = 0 for all t ∈ [0, 1]?

Question

Let f(t) be a continuous function and
f(t) = 0 for almost all t ∈ [0, 1].

Show that f(t) = 0 for all t ∈ [0, 1].

Answer 1

Since f(t) = 0 for almost all t ∈ [0, 1], it follows the set E = {x in (0,1) | f(t) <>0} has Lebesgue measure zero. So, E can't contain a nonempty interval, which means it has an empty interior which, in turn, means the set E' =[0,1] - E is dense in [0,1].

So, the function f and the function g(t) = 0 are both continuous in [0,1] and agree on the sett E', which is dense in [0,1]. This implies f = g and, therefore, f(t) = 0 for all ∈ [0, 1].

To see f = g, take a t in [0,1] and let t_n be any sequence in E' that converges to t. Since E' is dense in [0,1], this sequence exists. Since f and g are continuous, lim f(t_n) = f(t) and lim g(t_n) = g(t). Since f and g agree on E', f(t_n) and g(t_n) are the same sequence which, in virtue of the uniqueness of the limit, implies f(t) = g(t) = 0 for every t in [0, 1].

Answer 2

Let A={t∈(0,1);f(t) not equal to 0}. Since f is continuous, A is open. In A, choose any open interval (a,b); its measure is b-a. Then b-a<=measure(A). But since f(t) = 0 for almost all t ∈ [0, 1], f(t) = 0 for almost all t ∈ (0, 1), hence measure(A)=0, thus a=b. Hence the open set A contains only empty intervals, hence A is empty. Finally, since f is continuous and equals zero over (0,1), f(0)=f(1)=0 so f is zero throughout [0,1].


@Steiner: Your proof can improved if you note that since f is continuous, E is open, and the emptiness of its interior then means E is empty and E'=[0,1]. This will simplify your proof.

Answer 3

intuitively speaking, if there is a point t in [0, 1], f(t) <> 0 , then

there are inifinitely many points in [0, 1] whose image is different from 0 (since f is continuous) * in fact you can find an interval contained in (0, 1) whose image will be non zero in that case.

contradicting the given, that f(t) = 0 for almost all t.

Answer 4

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Answer 5

Proof is something like this:
let f(t) be continuous at p, for each epsilon[E] there exist delta[D] such that
f(p) - E < f(t) < f(p) + E for all x, p - D < x < p + D.

{(p-D,p+D)} covers [0,1]. Heine Borel says that there is a finite subcover.
Conclusion: for every q in [0,1}, choose the appropriate subinterval from the subcover to confirm continuity at q.