problems on Lebesgue integral 
1) Given a measure space, $f$ a non-negative measurable function and $A$ in the $\sigma$-algebra such that $\mu(A)=0$, prove that $\displaystyle\int_{A} f\;d\mu=0$ .

My try:
$$0\leq \int_{A}f\;d\mu=\int fI_{A}\;d\mu\leq \int \sup{f(A)}I_{A}\;d\mu=\sup{f(A)}\int I_{A}\;d\mu=\sup{f(A)}\mu(A)=0\;,$$
then $\displaystyle\int_{A}f\;d\mu=0\,.$

2) Given a measure space, let $f$ be an integrable function. Show that $\displaystyle \int_{\{|f|>t\}} |f|\;d\mu\rightarrow 0$ as $t\rightarrow\infty$ .

My try: since $f$ is integrable, $|f|\lt \infty$ almost everywhere, $\mu (\{|f|>t\}) = \mu(\Omega)- \mu(\{|f|\leq t\})$, and as $t\rightarrow\infty$,  $\mu (\{|f|>t\})<\epsilon$, then 
$$\int_{\{|f|>t\}} |f|\;d\mu=\int |f|I_{\{|f|>t\}}\;d\mu\leq\max\{|f|\}\int I_{\{|f|>t\}}=\max\{|f|\}\;\mu(\{|f|>t\})<\max\{|f|\}\;\epsilon\;.$$
Then it's done; I can take $\epsilon=\frac1t$.
Well, both demonstrations seem right for me. I will be thankful if someone tells me what's wrong on them.
Greetings.
 A: 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.
A: 1) I think a more natural argument would be to first prove the result in the case where $f$ is a simple function, then extend via the monotone convergence theorem. That being said, I like your solution, although something about it seems weird.
2) This argument fails if $\max|f|$ doesn't exists; i.e. if $\sup|f| = \infty$. I recommend proving by contradiction: If the result fails, then there is $\epsilon>0$ and a sequence $t_1 < t_2 < t_3 <\cdots$ with $t_i \to \infty$ such that $$\int_{\{|f|>t_{i}\}} |f| \, du > \epsilon$$ for all $i$. Using the monotone convergence theorem, you can then show that $$\sum^{\infty}_{i=k} \int_{\{t_i < |f| \leq t_{i+1}\}} |f|\, du = \int_{\{|f|>t_k\}}|f|\, du >\epsilon$$ for every $k$, so that sum on the left-hand side must be infinite (for each $k$), contrary to $f$'s integrability.
A: Solution to (2): I replaced t with n cause that's how I like it, haha. Let $E_n = \{ x: |f(x)|>n \}$. Note that $E_n \supset E_{n+1}$, we have a decreasing 
sequence of sets. And $\bigcap_n E_n = \{ x: |f(x)| = \infty \}$.
Now since $f$ is integrable, so is $|f|$. We define a measure $\lambda(E) = \int_E |f| d\mu$
for all $E$ in the $\sigma$-algebra.
Note that $\lambda(E_1) = \int_{E_1} 1 d\mu\ \leq \int_{E_1} |f| d\mu\ < \infty$. 
And since $\mu(\bigcap_n E_n) = 0, \lambda(\bigcap_n E_n) = 0$. Thus
\begin{equation*}
\lim_{n \to \infty} \int_{E_n} |f| d\mu\ =
\lim_{n \to \infty} \lambda(E_n) = \lambda(\bigcap_n E_n) = 0.
\end{equation*}
Remark: Since $f$ is integrable, so is $|f|$. Therefore if it so happens that $|f| = \infty$ at some set, that set has measure zero! So for this problem, we need not worry about it.
A: *

*First suppose $f\in L_1(\mu)$. The Markov-Chebyshev inequality implies that
$$\mu(|f|>n)\leq\frac{1}{n}\int_{\{|f|>n\}}|f|\,d\mu\leq\frac{1}{n}\int|f|\,d\mu\xrightarrow{n\rightarrow\infty}0$$
Notice that for any $n\in\mathbb{N}$, $\{|f|=\infty\}=\bigcap_m\{|f|>m\}\subset\{|f|>n\}$ and so $\mu(|f|=\infty)\leq\frac{1}{n}\int|f|\,d\mu\xrightarrow{n\rightarrow\infty}0$.

*Now that finiteness almost surely (a.s) of $f$ has been established, you can use the dominated convergence: Let $h_n=|f|\mathbb{1}_{\{|f|>n\}}$. Notice that $|h_n|\leq|f|$ and that $h_n\xrightarrow{n\rightarrow\infty}0$ $\mu$-a.s.

