The Real And Complex Number Systems

which implies that
h (x + t) − h (x)
t
≥ h m (x + t) − h m (x)
t
for all m. (*)
Since h and h m are increasing, we have h ′ and h ′ m exists a.e. on [a, b] . Hence,
by (*)
m∑
h ′ m (x) = |f nk (t) − g (t)| p ≤ h ′ (x) a.e. on [a, b]
k=1
which implies that
∞∑
|f nk (t) − g (t)| p exists a.e. on [a, b] .
k=1
So, f nk (t) → g (t) a.e. on [a, b] . In addition, f n → f on [a, b] . Then we
conclude that f = g a.e. on [a, b] . Since f and g are continuous on [a, b] , we
have
∫ b
a
|f − g| dx = 0
which implies that f = g on [a, b] . In particular, as p = 2, we have f = g.
Remark: (1) A property is said to hold almost everywhere on a set
S (written: a.e. on S) if it holds everywhere on S except for a set of measurer
zero. Also, see the textbook, pp 254.
(2) In this proof, we use the theorem which states: A monotonic function
h defined on [a, b] , then h is differentiable a.e. on [a, b] . The reader can
see the book, The reader can see the book, Measure and Integral (An
Introduction to Real Analysis) written by Richard L. Wheeden and
Antoni Zygmund, pp 113.
(3) There is another proof by using Fatou’s lemma: Let {f k } be a
measruable function defined on a measure set E. If f k ≥ φ a.e. on E and
φ ∈ L (E) , then ∫
∫
lim k ≤ lim inf f k .
E
k→∞ k→∞
E
Proof: It suffices to show that f nk (t) → g (t) a.e. on [a, b] . Since
l.i.m. n→∞ f n = g on [a, b] , and given ε > 0, there exists a n k such that
∫ b
a
|f nk − g| 2 dx < 1 2 k
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