Suppose, $S_{1}$, $S_{2}$ & $S_{3}$ are measurable subsets of $[0,1]$, each of measure $\dfrac{3}{4}$ such that the measure of $S_{1}\cup S_{2}\cup S_{3}$ is $1$. Then the measure of $S_{1}\cap S_{2}\cap S_{3}$ lies in :

(a) $\left[0,\dfrac{1}{16}\right]$

(b) $\left[\dfrac{1}{16},\dfrac{1}{8}\right]$

(c) $\left[\dfrac{1}{8},\dfrac{1}{4}\right]$

(d) $\left[\dfrac{1}{4},1\right]$.

We know that, $$m\left(S_{1}\cup S_{2}\cup S_{3}\right)=m\left(S_{1}\right)+m\left(S_{2}\right)+m\left(S_{3}\right)-m\left(S_{1}\cap S_{2}\right)-m\left(S_{2}\cap S_{3}\right)-m\left(S_{1}\cap S_{3}\right)+m\left(S_{1}\cap S_{2}\cap S_{3}\right).$$

From this relation how we can determine the required interval?

  • 1
    $\begingroup$ One possible hint: let $T_i = [0,1]\setminus S_i$. Then each $T_i$ has measure $\frac14$ and $T_1 \cap T_2 \cap T_3$ has measure $0$. What interval must the measure of $T_1 \cup T_2 \cup T_3$ lie in? $\endgroup$ Jan 8, 2015 at 8:23
  • $\begingroup$ One (slightly unfair) approach would be to just pick one example of sets as in the hypothesis and to check in which interval you land. Since the intervals are (almost) disjoint, this can answer the question. But this assumes of course that at most one answer is correct. $\endgroup$
    – PhoemueX
    Jan 8, 2015 at 9:16

1 Answer 1


Your relation there is incorrect. You need $\ldots + m(S_1\cap S_2 \cap S_3)$ rather than the minus you've got.

Plugging in your numbers you get $$1 - 9/4 + m(S_1\cap S_2) + m(S_1\cap S_3) + m(S_2\cap S_3) = m(S_1\cap S_2 \cap S_3).$$ Now, $1/2 \leq m(S_i\cap S_j) \leq 3/4$ and plugging that in we have $$ - 5/4 + 3/2 \leq m(S_1\cap S_2 \cap S_3) \leq 1$$ which is (d).

This is actually a pretty loose bound. There's no way to get close to 1.

  • $\begingroup$ Why $1/2\le m(S_{i}\cap S_{j})$? $\endgroup$
    – Empty
    Jan 30, 2015 at 5:10
  • $\begingroup$ Sorry..$-$ , $+$ was the typing mistake..see editing .... $\endgroup$
    – Empty
    Jan 30, 2015 at 5:12
  • $\begingroup$ Let $A = m(S_1\backslash S_2), B = m(S_2\backslash S_1), C = m(S_1\cap S_2)$ and $D = m([0, 1] \backslash (S_1 \cup S_2)).$ $A+B+C+D = 1$, $A+C = B+C = 3/4$. So, $C-D = 1/2$, and $D$ is positive. The lower bound follows. $\endgroup$
    – user24142
    Jan 30, 2015 at 5:26

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