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Let $f:\mathbb{R} \to \mathbb{R}$ be defined by

$f(x)=\left\{\begin{array}{lllll} x+5 & \quad\text{if }x<-1 \\\ 2 & \quad\text{if } -1\leq\alpha<0\\\ x^2 & \quad\text{if } x \geq 0\end{array}\right.$

Show that $f$ is measurable function.

we know that f is measurable if

(i) It's domain is measurable, and

(ii) for any a belong to R one of the following is satisfy

${x | f(x) > a }$ is measurable

${x | f(x) >= a }$ is measurable

${x | f(x) < a }$ is measurable

${x | f(x) <= a }$ is measurable

but, I can't find any a such that.

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  • $\begingroup$ what is $f^{-1}((t,\infty))$ for variuos choice of $t$? $\endgroup$
    – GA316
    Nov 7, 2013 at 8:20
  • $\begingroup$ Construct, for example, the set $E_f (a)= \{x|\;\; f(x) < a \}$ for any $a\in\mathbb{R}.$ Is it measurable? $\endgroup$ Nov 7, 2013 at 8:24

3 Answers 3

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First, plot the function and slice the plan for every relevant value of $\alpha$ (it will appear evident, once you plotted the function).

It takes a bit of practice, but you then compute the inverse image for every relevant value of $\alpha$, and find:

$\mathbb{R}(f<\alpha)=\left\{\begin{array}{lllll} \left[-\infty,\alpha-5\right[ & \quad\text{if }\alpha <0\\\left[-\infty,-5\right]\cup\{0\} & \quad\text{if }\alpha=0\\\left[-\infty,\alpha-5\right]\cup\left[0,\sqrt{\alpha}\right] & \quad\text{if } 0<\alpha<2\\\left[-\infty,\alpha-5\right]\cup \left[-1,\sqrt{\alpha}\right] & \quad\text{if } 2\leq\alpha<4\\\left[-\infty,\sqrt{\alpha}\right] & \quad\text{if } 4\leq\alpha\end{array}\right.$

Every set of the right is measurable, so the function $f$ is measurable.

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  • $\begingroup$ Can you please show the plot ? $\endgroup$
    – The Doctor
    Feb 18, 2017 at 11:30
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Let $a$ be a real number, and $B=\{x\in\mathbb R : f(x)<a\}$.

Let $I_1=(-\infty,-1)$, $I_2=[-1,0)$ and $I_3=[0,+\infty)$ and $B_k=B\cap I_k$ for $k\in\{1,2,3\}$.

$B$ is the (disjoint) union of the $B_k$ subsets.

Then $x$ is in $B_1$ iff $x\in I_1$ and $x+5<a$, i.e. $x\in(-\infty,a-5)\cap I_1$.

Hence, $B_1=(-\infty,a-5)\cap I_1$ is a measurable set since it's the intersection of two intervals.

Now to $B_2$ : $x$ is in $B_2$ iff $x\in I_2$ and $a=2$.

So $B_2=I_2$ if $a=2$ and $B_2=\emptyset$ if $a\neq 2$. It's a measurable set in both cases.

And finally, $x$ is in $B_3$ iff $x\in I_3$ and $x^2<a$.

So $B_3=I_3\cap (-\sqrt{a},\sqrt{a})$ if $a>0$ and $B_3=\emptyset$ if $a\leqslant 0$. Again, it's a measurable set in both cases.

Finally, $B$ is measurable as a (finite) union of measurable sets.


IMO, the easiest way to deal with such questions is to write

$$f(x)=(x+5)\chi_{(-\infty,-1)}+2\chi_{[-1,0)}+x^2\chi_{[0,+\infty)}$$

  • Intervals are measurable, hence each $\chi_I$ is a measurable function.
  • Polynomial functions are continuous hence measurable.
  • Sums and products of measurable functions are measurable.

Hence $f$ is measurable.

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  • $\begingroup$ the last way is very easy but i need the first $\endgroup$
    – Hamada Al
    Nov 7, 2013 at 8:48
  • $\begingroup$ You really can't do this alone? I'll add a few more lines to my answer. $\endgroup$ Nov 7, 2013 at 10:48
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what is $f^{-1}((t,\infty))$ if i) $t \ge 4$, ii)$ 2 \lt t \lt 4$ , iii) $t = 2 $ iv) $0 \le 2 \lt 2$ and v) $t \lt 0$

prove that these sets are measurable

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  • $\begingroup$ i don,t know that i am so sorry $\endgroup$
    – Hamada Al
    Nov 7, 2013 at 8:27
  • $\begingroup$ draw the graph of the function and find the set $f ^{-1}((t,\infty))$ for those values of $t$. $\endgroup$
    – GA316
    Nov 7, 2013 at 8:28

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