Metric for the Space of Measurable Functions

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46MATHEMATICS: A. C. BERRY

456

PROC. N. A. S.

A METRIC FOR THE SPACE OF MEASURABLE FUNCTIONS By ANDREW C. BERRY* DEPARTMENT OF MATHEMATICS, PRINCETON UNIVERSITY

Communicated June 30, 1931

Fr6chet' cites a metric for the space of measurable functions and shows that the space is complete with respect to this metric. He notes, however, that the metric does not satisfy the triangle inequality, and is not homogeneous. The present paper furnishes a metric which (Theorem 4) satisfies the triangle inequality, which (Theorem 5) possesses quasi-homogeneity, and for which (Theorem 6) the space is complete. DEFINITION. A complex-valued, Lebesgue measurable function f(x) of the single real variable x, - co < x < co, shall be said to be an element of the space M if there exists, as a finite number, I(f) = greatest lower bound of real numbers e such that meas

I I f(x)

> e

}


A I ?s -a } > X-6, for each a >O. meas {f(x) I > Proof of Necessity.-Let X = I(f). By definition, the second requirement is met. Again by definition, I meas { jf(x) >1(f) + } _ I(f) + eforeache > 0. As e o- 0, the set { f(x) > I(f) + e } expands toward the limit set { | f(x) > I(f) } and so lim meas = meas lim. Thus, meas { jf(x) > I(f) } < I(f), and the first requirement is met. Proof of Sufficiency.-Let X satisfy the two requirements. Then, by definition, - 6 0, which is possible only if X = I(f). THEOREM 2. 1(f) > 0. Proof.-The inequality meas { | f(x) > e } _ e can be satisfied only by non-negative values of e since measure is a non-negative quantity. THEOREM 3. A necessary and sufficient condition that l(f) = 0 is that f(x) = 0 almost everywhere. Proof.-By Theorems 1 and 2, a necessary and sufficient condition

I

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MATHEMATICS: A. C. BERRY

VOL. 17, 1931

that l(f) = 0 is that meas I f(x) > 0 } = 0, which is equivalent to the statement that f(x) = 0 almost everywhere. THEOREM 4. l(f + g) _ I(f) + I(g). Proof.-If x is such that

f(x) + g(x)j

>

I(f) + 1(g),

then x must be such that at least one of the inequalities

f(x) > I(f),

g(x) > I(g)

is satisfied. Hence, meas jf(x) + g(x) | > I(f) + 1(g) } _ meas If(x) | > I(f) } + meas { g(x) > I(g) }.1(f) + (g), the last inequality being a consequence of Theorem 1. It follows, by definition, that l(f + g) 1, then l(af) g I a Il(f). Proof.-By Theorem 1, when a > 1, meas

(f) } = meas { jf(x)

{|af(x) > a

> 1(f) }

_1(f) .

a

|(f)-

Hence, by definition, l(af) < a 1(f) when a _ 1. THEOREM 6.2 If { f"(x) } is a sequence of elements of M such that Jim l(fm - fM) = 0, then there exists an element f(x) of M such that

lim l(f - f,) = 0.

Proof.-We can determine a sub-sequence { f,*(x) } such that l(fm -f) < 2-k when m _ nk and n _ nk, k = 1, 2,. In particular, by Theorem 1, meas

lfnk+l(x) -fnk(x)

{

>

2-k }