Suppose that every point of an interval X = [ 0 , 1 ] lies in at least k subintervals A i ⊂ X, i = 1 , . . . , n. Show that k ≤ ∑ i = 1 n | A i | where | A i | denotes the length of A i .My thoughts:Let's consider k copies of X. We will denote j-th copy as X j .If we would be able to cover them with { A i } i = 1 n then we're done.(it is also enough to left uncovered the set of measure zero in every copy).Let's fix an arbitrary point x ∈ X 1 . There exists 1 ≤ i 1 &lt; . . . &lt; i k ≤ n such that x ∈ A i j , j = 1 , . . . , k. Then we will use A i j to cover everything it can on the copy X j . Now, let's choose another point x ∈ X 1 and repeat the process.There is a hard part in this proof. I can't understand why it can not be the case when we used all our A i but didn't fully cover every copy X j (or didn't left uncovered the set of measure zero in every copy).

Question

Suppose that every point of an interval   X  =  [  0  ,  1  ] lies in at least   k subintervals       A    i    ⊂  X,   i  =  1  ,  .  .  .  ,  n. Show that  k  ≤      ∑          i      =      1        n        |        A    i        |  where       |        A    i        |   denotes the length of       A    i  .My thoughts:Let&#039;s consider   k copies of   X. We will denote j-th copy as       X    j  .If we would be able to cover them with   {      A    i        }          i      =      1        n   then we&#039;re done.(it is also enough to left uncovered the set of measure zero in every copy).Let&#039;s fix an arbitrary point   x  ∈      X    1  . There exists   1  ≤      i    1    &amp;lt;  .  .  .  &amp;lt;      i    k    ≤  n such that   x  ∈      A                  i        j              ,  j  =  1  ,  .  .  .  ,  k. Then we will use       A                  i        j             to cover everything it can on the copy       X    j  . Now, let&#039;s choose another point   x  ∈      X    1   and repeat the process.There is a hard part in this proof. I can&#039;t understand why it can not be the case when we used all our       A    i   but didn&#039;t fully cover every copy       X    j  (or didn&#039;t left uncovered the set of measure zero in every copy).

Harold Cantrell · Accepted Answer

This is easy:       ∑    i        χ                  A        i              (  x  )  ≥  k for each   x  ∈  X. [Each term is 0 or 1 and at least   k terms are equal to 1]. Integrating we get   k  ≤  ∑      |        A    i        |  ,

Semaj Christian · Accepted Answer

It&#039;s worth mentioning that your solution can be easily extended to an abstract measure space   (  X  ,  Σ  ,  μ  )

Suppose that every point of an interval X = [ 0 , 1 ] lies in at least k

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