Consider $f(x) = -1$ for all $x \in \mathbb{R}$.
People A say: $\int f dx = -\infty$.
People B say: $\int f dx$ does not exist!
Everyone agrees that: $-\infty < \infty$.
So it is reasonable for people A to say:
$$ \int f dx = -\infty < \infty$$
On the other hand, it is reasonable for people B to say: $\int f dx$ is not less than $\infty$ because $\int f dx$ does not exist, and something that does not exist cannot be compared in an inequality statement!
Convention for People A
Let $(A, \mathcal{F}, \mu)$ be a measure space triplet. If $f:A\rightarrow\mathbb{R}$ then define
\begin{align}
f^+(x) &= \max[f(x),0] \quad \forall x \in A \\
f^-(x) &= -\min[f(x),0] \quad \forall x \in A
\end{align}
Then $f^+$ and $f^-$ are nonnegative functions and
\begin{align}
f(x) &= f^+(x) - f^-(x) \quad \forall x \in A\\
|f(x)| &= f^+(x) + f^-(x) \quad \forall x \in A
\end{align}
It can be shown that if $f$ is measurable then $f^+, f^-, |f|$ are nonnegative and measurable.
If $f$ is a nonnegative and measurable function then $\int fd\mu$ is always defined (possibly equal to $\infty$). Also it satisfies:
$$ \int f d\mu = \lim_{M\rightarrow\infty} \int_0^M \mu(\{x \in A : f(x)\geq t\})dt $$
If $f$ is a measurable function (possibly taking negative values) then we define
$$ \int f d\mu = \int f^+ d\mu - \int f^- d\mu$$
whenever the right-hand-side avoids the undefined case of $\infty - \infty$. That is, $\int f d\mu$ is defined if and only if either $\int f^+d\mu <\infty$ or $\int f^-d\mu < \infty$.
With this definition we observe that
$$ \int |f|d\mu = \int f^+d\mu + \int f^-d\mu$$
and $\int |f|d\mu <\infty$ if and only if $\int f^+d\mu<\infty$ and $\int f^-d\mu<\infty$.
Convention for People B
Everything is the same, except that people B require all integrals $\int f d\mu$ to have finite values. So they say that $\int f d\mu$ is defined if and only if $\int f^+d\mu<\infty$ and $\int f^- d\mu<\infty$. That is, they say that $\int f d\mu$ is defined if and only if $\int |f|d\mu<\infty$.
The advantage of this convention is that it removes the cases when $\int f^+d\mu$ or $\int f^-d\mu$ are infinite, so there is often less work to do (we do not have to consider so many cases when proving things). However, people B routinely use the people A language. For example, people B will routinely use the Borel-Cantelli lemma by saying things like
- "If $\sum_{n=1}^{\infty} P[A_i]=\infty$ then..."
but of course, strictly speaking, the equation $\sum_{n=1}^{\infty} P[A_i]=\infty$ is not allowed to exist as an equation under the people B convention: If the equation is true then the left-hand-side does not exist and so the equation is meaningless!
Now this Borel-Cantelli example uses a sum instead of an integral, but the people B convention for sums is (usually) the same as for integrals, and in fact a sum can be viewed as an integral under a certain measure.
One can also observe that the people B convention is problematic because it implies that the inequality $\int f d\mu < \infty$
is either meaningless or trivially true: If the integral $\int f d\mu$ is allowed to exist then (by the people B convention) it must be finite and so there is no reason to write such an inequality (it is trivially true). On the other hand if $\int f d\mu$ does not exist then we are not allowed to pretend that it exists in the inequality $\int fd\mu < \infty$. People B must live with these (minor) contradictions.
If you want the best of both worlds, being free of contradictions and also doing less work, you can just write the preamble "Suppose $\int |f|d\mu <\infty$" before your analysis of integrals of $f$. Under the condition $\int |f|d\mu <\infty$, people A and B can co-exist happily.
Best Answer
Edit: I missed some things in my first answer and updated my answer accordingly.
Edit 2: Not only that I missed some things in my first answer, I missed the point completely. I leave the answer here for instructional purposes. Tasadduk's answer is definitely the way to go about 2).
I agree with mac that the first solution is almost ok, but some extra care has to be taken (the supremum might be infinite and even if one could argue that $\infty \cdot 0 = 0$ in measure theory, a clear cut argument is certainly preferable to a convention).
I would first prove that for a non-negative and bounded functions $f$ we have $\int_{A} f \,d\mu = 0$, for this you can use your argument. The monotone convergence theorem then implies that it holds for all non-negative measurable functions, simply by approximating $f$ by $f_{n} = \min{\{f,n\}}$ from below.
Remark. The equality $\int_{A} f = 0$ holds true for all measurable $f$, because a general $f$ can be written as $f = f_+ - f_-$ with $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$ and use the definition $\int_{A} f\,d\mu = \int_{A} f_+ \,d\mu - \int_{A} f_-\,d\mu$ provided at least one of the integrals of the right hand side is finite.
As mac also pointed out, there are quite a few problems in the second proof and I don't see how to save it using your idea.
Here's how I would do it. Let $A_{t} = \{|f| \gt t\}$ and note that for $s \lt t$ we have $A_{s} \supset A_t$. Define $$a_{t} = \int_{\{|f| \gt t\}} |f|\,d\mu = \int_{A_t} |f|\,d\mu.$$ Since $|f| \geq 0$ and $A_{s} \supset A_{t}$ we have $a_{s} \geq a_{t}$ for $s \leq t$. We want to show that $a_{t} \to 0$ as $t \to \infty$. By monotonicity of $t \mapsto a_{t}$, it suffices to show that $a_{n} \to 0$ as $n \to \infty$ runs through the natural numbers. Since $a_{n} \geq 0$ and $a_{n}$ is monotonically decreasing, the limit $a = \lim_{n \to \infty} a_{n}$ exists, and we want to show that $$ a = 0.$$
Now let $f_{n}(x) = \min{\{|f(x)|,n\}}$ and note that $f_{n} \to |f|$ pointwise (and monotonically). By the monotone convergence theorem (or the dominated convergence theorem, applicable since $f_{n} \leq |f|$ and $|f|$ is integrable, if you prefer) we have
$$ \int |f|\,d\mu = \lim_{n \to \infty} \int f_{n}\,d\mu.$$
On the other hand $n \leq |f|$ on $A_{n}$ and $0 \leq f_{n} \leq |f|$ on $\Omega$ imply $$ \int f_{n}\,d\mu = \int_{A_{n}} n \,d\mu + \int_{\Omega \smallsetminus A_{n}} f_{n}\,d\mu \leq \int_{A_{n}} |f|\, d\mu + \int |f| \,d\mu = a_{n} + \int |f|\,d\mu.$$
Passing to the limit on both sides of the estimate $\int f_{n} \leq a_{n} + \int |f|$ we get $$\int |f|\,d\mu = a + \int |f|\,d\mu$$ and as $\int |f|\,d\mu \lt \infty$ we conclude $a = 0$, as we wanted.