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.

  • $\begingroup$ 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


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.

  • $\begingroup$ Can you please show the plot ? $\endgroup$
    – The Doctor
    Feb 18, 2017 at 11:30

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


  • 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.

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.