P-Seminorm of Function Zero iff A.E. Zero

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \closedint 1 \infty$.

Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space.

Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.

Then:

$\norm f_p = 0$ if and only if $f = 0$ $\mu$-almost everywhere

where $\norm \cdot_p$ is the $p$-seminorm.

Proof

Case 1: $1 \le p < \infty$

We have that:

$\norm f_p = 0$ if and only if $\norm f_p^p = 0$.

That is, $\norm f_p = 0$ if and only if:

$\ds \int \size f^p \rd \mu = 0$

From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, this is equivalent to:

$\size f^p = 0$ almost everywhere.

From Pointwise Exponentiation preserves A.E. Equality, we have:

$\size f = 0$ almost everywhere.

So:

$f = 0$ almost everywhere.

$\Box$

Case 2: $p = \infty$

We have that:

$\norm f_\infty = 0$

if and only if:

$\inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0} = 0$

From the definition of infimum, for each $\epsilon$ there exists $c < \epsilon$ such that:

$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$

We show that it follows that:

$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$

for all $c > 0$.

Aiming for a contradiction, suppose not, and there exists $M > 0$ such that:

$\map \mu {\set {x \in X : \size {\map f x} \ge M} } > 0$

Then, from Survival Function is Decreasing, we have:

$\map \mu {\set {x \in X : \size {\map f x} \ge c} } > 0$

for all $c \le M$.

But we have shown that there must exist $c \le M$ such that:

$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$

which is a contradiction.

So:

$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$

for all $c > 0$.

Then, we have:

\(\ds \map \mu {\set {x \in X : \size {\map f x} > 0} }\)

\(=\)

\(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty \set {x \in X : \size {\map f x} \ge \frac 1 n} }\)

\(\ds \)

\(=\)

\(\ds 0\)

Null Sets Closed under Countable Union

Then, if $\map f x \ne 0$ we have:

$x \in \set {x \in X : \size {\map f x} > 0}$

which is a $\mu$-null set.

So:

$f = 0$ $\mu$-almost everywhere.

$\blacksquare$