$f=0$ a.e. iff $\int_E f=0$ for every measurable $E$

Show that $f=0$ a.e. iff $\int_E f = 0$ for every measurable set $E$.

Solution

Let $f$ be a real-valued function in $L^1(\mu)$, so that every integral $\int_E f$ is defined.

1. ($\Rightarrow$) If $f = 0$ a.e. then $\int_E f = 0$ for every measurable $E$.

Functions equal a.e. have equal integrals over $E$, and $\int_E 0 = 0$.

2. ($\Leftarrow$) If $\int_E f = 0$ for every measurable $E$, then $f = 0$ a.e.

2.1. $\mu\theset{f > 0} = 0$.

Apply the hypothesis to $E \coloneqq \theset{f > 0}$. On $E$, $f = f^+ > 0$ and $f^- = 0$, so $0 = \int_E f = \int_E f^+$. A nonnegative measurable function with zero integral is zero a.e., so $f = 0$ a.e. on $E$; since $f > 0$ on $E$, $\mu(E) = 0$.

2.2. $\mu\theset{f < 0} = 0$.

Apply the hypothesis to $E \coloneqq \theset{f < 0}$: $\int_E f = -\int_E f^- = 0$, so $f^- = 0$ a.e. on $E$; since $f^- > 0$ on $E$, $\mu(E) = 0$.

2.3. Q.E.D.

$\theset{f \neq 0} = \theset{f > 0} \cup \theset{f < 0}$ has measure $0$ by steps 1 and 2.

3. Q.E.D.

Steps 1 and 2 are the two implications.

Source Collections

Unsorted real analysis problems, problem 7