Let f ∈ L 1 ( R 2 ) with respect to the Lebesgue measure m × m on R 2 . Prove that if ∬ R 2 f ( x , y ) d x d y = 0 ,then there exits a square S a , b = { ( x , y ) ∣ a ≤ x ≤ a + 1 , b ≤ y ≤ b + 1 }, such that ∬ S a , b f ( x , y ) d x d y = 0.I tried to show that the integral ∬ [ a , a + x ] × [ b , b + y ] f ( s , t ) d s d tis absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.Is there any theorem for the two-dimensional case?

Question

Let   f  ∈      L    1    (            R        2    ) with respect to the Lebesgue measure   m  ×  m on             R        2  . Prove that if      ∬                            R                2              f  (  x  ,  y  )    d  x    d  y  =  0  ,then there exits a square       S          a      ,      b        =  {  (  x  ,  y  )  ∣  a  ≤  x  ≤  a  +  1  ,  b  ≤  y  ≤  b  +  1  }, such that      ∬                  S                  a          ,          b                      f  (  x  ,  y  )    d  x    d  y  =  0.I tried to show that the integral      ∬          [      a      ,      a      +      x      ]      ×      [      b      ,      b      +      y      ]        f  (  s  ,  t  )    d  s    d  tis absolutely continuous by Fubini&#039;s Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.Is there any theorem for the two-dimensional case?

Zackary Galloway · Accepted Answer

Let   g  (  a  ,  b  )  =      ∬                  S                  a          b                      f  (  s  ,  t  )    d  s    d  t.   g is a continuous real valued function on             R              2      . If it is never zero it is always positive or always negative. [ Because its range is connected in       R  ]. If it is always positive then       ∬                            R                          2                      f  (  s  ,  t  )    d  s    d  t  &amp;gt;  0 because this integral is the sum of integrlas over       S          n      m       as   n  ,  m vary over all integers.

Let f in L^1(R^2) with respect to the Lebesgue measure m*m on R^2. Prove that if int int_(R^2) f(x,y)dxdy=0

Answered question

Answer & Explanation

New Questions in Calculus 2