Let $f:X \to \mathop{R}^*$ be finite for almost all $x$. Suppose that $X$ is a finite measure. Prove that for each $\varepsilon >0$, there exists an $A$ s.t. $\mu (X-A)<\varepsilon$ and $f$ is bounded on $A$. Note: $\mathop{R}^*$ is the extended real line.
My attempt:
Take $A:=\{f<1/\varepsilon \}$. Clearly, $\mu (X-A)=\mu\{f \geq 1/\varepsilon \}$.
Now, $\mu (X-A)\to \mu \{f=\infty\}=0$ as $\varepsilon \to 0^+ $.
So, somehow I can get that for small enough $\varepsilon$, $\mu (X-A)< \varepsilon$.
Is my attempt true?
Is there an example that shows that $X-A$ fails to be null?
your solution is basically correct, but it needs to be more "formal". Basically what you do is to use continuity from above, that is, given a measure $\mu$ and a sequence of measurable sets $A_0\supseteq A_1\supseteq \cdots$, if $\mu(A_0)<\infty$ (this condition is the reason for asking $\mu(X)<\infty$), then $$\mu(\bigcap_{n=0}^\infty A_n)=\lim_n\mu(A_n)$$ Now consider the sequence $A_n=\{|f|\geqslant n\}$ and notice that $A_0=X$ and the sequence $\{A_n:n\in\mathbb{N}\}$ is a ${descending}$ sequence of measurable sets, so $$0=\mu(\{f=\pm\infty\})=\mu(\bigcap_{n=0}^\infty A_n)=\lim_n\mu(A_n)$$ at this point use the definition of limit, for each $\varepsilon>0$ you can find a $N$ such that $\mu(A_n)<\varepsilon\forall n>N$, so for each $n>N$, $A=X-A_n$ is the set you are looking for, since $\mu(X-A)=\mu(A_n)<\varepsilon$ and $|f(z)|<n$ for each $z\in A$. To show how the condition $\mu(X)<\infty$ is necessary, consider $X=\mathbb{R}$ and $f=|x|$, then $f$ is $everywhere$ finite, but if $f$ is bounded in $A$, then $A$ itself must be bounded (since $f$ being bounded on $A$ means there is a $b$ positive real number s.t. $|z|<b\forall z\in A$), but if $|z|<b\forall z\in A\rightarrow [b,\infty[\subseteq \mathbb{R}-A\rightarrow \mu(\mathbb{R}-A)=\infty$