※ 引述《kemowu (小展)》之銘言：
: Suppose ∫ f(y) dy = 0 for every subinterval I of R. Show that f = 0 a.e.
: I
: on R.
For any finitely measurable set A, there corresponds a descending sequence
of open sets O_n, each containg A, such that the intersection of all O_n
differ only by a measure zero set from A. Let χ_n be the characteristic
function of O_n, χ be that of A. Since the integral of f vanishes on each open
interval, so does the integral of fχ_n. Hnece by Lebesgue dominated
convergence theorem, ∫f χ dx = ∫_A f dx =0. Namely, the integral of f on
each finitely measurable set equals zero, which implies the same property of f
on each measurable set.
For all positive a define E(a) := {x｜f(x)>a}.
0 =∫ f dx ≧ a m(E(a)).
E(a)
Hence m(E(a)) = 0 for all positive a. Consequently f(x)≦0 a.e.. The reversed
inequality holds similarly.
Another proof:
If we take Mikusinski formulation to Daniell integral as our definition
of integration, f is integrable ( on (a,b) ) iff
there exists a sequence of step functions s_n such that
(i) ∫|s_m-s_n| dx approaches zero as m and n tend to infinity,
(ii) s_n tends to f a.e.,
whence ∫f dx is defined by lim ∫s_n dx.
n→∞
Hence
│∫s_n dx│≦ ∫│s_m-s_n│dx + │∫ s_m dx│
I I I
t_n
Put s_n = Σ c χ , where χ is the charateristic function of
k=1 k,n k,n k,n an interval I_{k,n}.
Subsequently,
t_n t_n
Σ│∫ s_n dx│ = ∫│s_n│ dx ≦∫│s_m-s_n│dx + │Σ ∫ s_m dx│.
k=1 I_{k,n} k=1 I_{k,n}
Given ε>0, there correspond a large N such that ∫|s_m-s_n| dx < ε
whenever m > n > N. Fix n > N, and let m approaches infinity:
∫│s_n│ dx < ε, for all n > N.
Hence for all positive a, m({x│|s_n(x)| > a, for some n > N }) ≦ ε/a,
which implies m({x│|f(x)| ≧ a }) ≦ ε/a. Since ε was arbitrary,
m({ x | |f(x)| ≧ a }) = 0.
基本上提到 interval 就可以想想上面第二種證明一開始列的那個性質,
如果從 measure formulation 來看這個性質可以當作 Egoroff theorem 的一種變形,
第一種方法基本上就是把這件事的證明跑過一遍, 大概是這樣
※ 編輯: ppia 來自: 114.32.4.99 (08/18 20:11)
※ 編輯: ppia 來自: 114.32.4.99 (08/18 20:12)