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:


 * $\norm f_p = 0$ $f = 0$ $\mu$-almost everywhere

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

Case 1: $1 \le p < \infty$
We have that:


 * $\norm f_p = 0$ $\norm f_p^p = 0$.

That is, $\norm f_p = 0$ :


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

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


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

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


 * $\size f = 0$ almost everywhere.

So:


 * $f = 0$ almost everywhere.

Case 2: $p = \infty$
We have that:


 * $\norm f_\infty = 0$




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

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 $c > 0$.

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:


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

contradiction.

So:


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

for all $c > 0$.

Then, we have:

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.