$\int_\Omega f d\mu = 0 $ if and only if $f(x)=0$ almost everywhere

Question

can someone give me a hint on what kind of theorem/definition I should make use of to solve this?

Let $(\Omega,\mathfrak A, \mu)$ be a measure space and $f:\Omega \to \mathbb R$ a non-negative measurable function. Show that $$\int_\Omega f d\mu = 0 $$ if and only if $$f(x)=0$$ a.e.

Answer 1

One side:

Let it be that the nonnegative $f$ is not $0$ a.e.

That means exactly that a measurable set $A$ exists with $\mu A>0$ and $f(a)>0$ for $a\in A$.

Defining $A_n:=\{a\in A\mid f(a)\geq\frac1{n}\}$ we have $A=\bigcup_{n=1}^{\infty} A_n$ and consequently $\mu A_n>0$ for $n$ large enough (since $\mu A_n\to\mu A$ in this situation).

Then $f\geq\frac1n1_{A_n}$ leads to $\int fd\mu\geq\int\frac1{n}1_{A_n}d\mu=\frac1{n}\mu A_n>0$.

Other side:

If $f=0$ a.e. then for every stepfunction (i.e. measurable function taking only finite nonnegative values) $g=\sum_{i=1}^n c_n1_{B_n}\leq f$ we will have have $\mu B_i=0$ for $i=1,\dots,n$ and consequently $\int gd\mu=0$.

By definition $\int fd\mu$ is the supremum of such values $\int gd\mu$ so evidently $\int fd\mu=0$

Answer 2

Hint: Suppose there is some measurable set $E$ with $\mu(E)>0$ such that $f>0$ on $E$. Notice that $f\geq f\chi_E$ where $\chi_E$ is the indicator function of $E$. It follows that $$\int_{\Omega}f\mathrm{d}\mu\geq \int_{\Omega}f\chi_E\mathrm{d}\mu>0.$$

Explain this last inequality in detail. Can you do the other implication yourself?

Answer 3

Sketch:

(1) Suppose not, then there is a positive measure subset $A$ where $f\not=0$.

(2) Let $A_n$ be the set of points where $f>\frac{1}{n}$.

(3) Observe that $A_n\subseteq A_{n+1}$ and $\bigcup A_n=A$.

(4) Conclude that some $A_n$ has positive measure using the properties of measures.

(5) Conclude that $\int f d\mu\geq\frac{1}{n}\mu(A_n)>0$.

Answer 4

If @$f(x) = 0$ a.e. is clear that $\int_{\Omega} fd\mu = 0$.

Now, suppose that $f$ is non-negative, $\int_{\Omega} fd\mu = 0$, but $f$ is not $0$ a.e. It means that there is $A$ such that $\mu(A) >0$ and $f(x) \neq 0, \forall x \in A$.

Then,

$\int_{\Omega} f d\mu \geq \int_{A} f d\mu > 0$. Contradiction!