Let O = { ( x 1 , x 2 ) ∈ R 2 : 0 &lt; x 1 = x 2 &lt; 1 }, I = { N ∈ B : λ ( ( 0 , 1 ) ∩ N ) = 0 }, where λ is the Lebesgue measure and B is the Borel sigma algebra.Furthermore let μ ( A ) = { 0 ∃ A 1 , A 2 ∈ I , and ∃ B 1 , B 2 ∈ B , so A ⊆ ( A 1 × B 1 ) ∪ ( A 2 × B 2 ) 1 otherwise Show that μ ( O ) = 1.My thought is a contradiction argument. If μ ( O ) = 0, then there must exist N 1 , N 2 ∈ I and B1,B2∈B such that O ⊆ ( N 1 × B 1 ) ∪ ( N 2 × B 2 ). But since O is a line from the point (0,0) to the point (1,1), there cannot be a set with Lebesguemeasure 0 on the interval (0,1) on either axis. There is therefore no way to "describe" the line, because it goes from 0 to 1 on both axis, and therefore there does not exist N 1 , N 2 ∈ I so O ⊆ ( N 1 × B 1 ) ∪ ( N 2 × B 2 )My problem is that I understand why is the measure μ ( O ) = 1, I just can't seem to find the right proof technique.

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Let   O  =  {  (      x    1    ,      x    2    )  ∈            R        2      :    0  &amp;lt;      x    1    =      x    2    &amp;lt;  1  },   I  =  {  N  ∈      B      :    λ  (  (  0  ,  1  )  ∩  N  )  =  0  }, where   λ is the Lebesgue measure and       B   is the Borel sigma algebra.Furthermore let   μ  (  A  )  =      {                            0                          ∃                      A            1                    ,                      A            2                    ∈          I          ,                     and           ∃                      B            1                    ,                      B            2                    ∈                      B                    ,                     so           A          ⊆          (                      A            1                    ×                      B            1                    )          ∪          (                      A            2                    ×                      B            2                    )                                      1                          otherwise                        Show that   μ  (  O  )  =  1.My thought is a contradiction argument. If   μ  (  O  )  =  0, then there must exist       N    1    ,      N    2    ∈  I and B1,B2∈B such that   O  ⊆  (      N    1    ×      B    1    )  ∪  (      N    2    ×      B    2    ). But since   O is a line from the point (0,0) to the point (1,1), there cannot be a set with Lebesguemeasure 0 on the interval (0,1) on either axis. There is therefore no way to &quot;describe&quot; the line, because it goes from 0 to 1 on both axis, and therefore there does not exist       N    1    ,      N    2    ∈  I so   O  ⊆  (      N    1    ×      B    1    )  ∪  (      N    2    ×      B    2    )My problem is that I understand why is the measure   μ  (  O  )  =  1, I just can&#039;t seem to find the right proof technique.

sleuteleni7 · Accepted Answer

Your approach is fine. Assuming that   O  ⊆  (      N    1    ×      B    1    )  ∪  (      N    2    ×      B    2    ), we get   (  0  ,  1  )  ⊆      N    1    ∪      N    2  . But   λ  (  (      N    1    ∪      N    2    )  ∩  (  0  ,  1  )  )  =  0  &amp;lt;  λ  (  (  0  ,  1  )  ), which gives the required contradiction.

Let O = <mo fence="false" stretchy="false">{ ( x 1 </msub> ,

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