Given the step function g ( x , y ) = { 1 x 2 0 &lt; y &lt; x &lt; 1 − 1 y 2 0 &lt; x &lt; y &lt; 1 0 otherwise I want to show that ∫ ( 0 , 1 ) ∫ ( 0 , 1 ) g ( x , y ) d λ ( x ) d λ ( y ) ≠ ∫ ( 0 , 1 ) ∫ ( 0 , 1 ) g ( x , y ) d λ ( y ) d λ ( x )However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as ∫ 0 x ∫ 0 1 1 x 2 d x d y − ∫ 0 1 ∫ 0 y 1 y 2 d x d yI also considered using limits such as following one, but again I end up with divergence. lim k → ∞ ∫ ( 1 k , 1 ) lim k → ∞ ∫ ( 1 k , 1 ) g ( x , y ) d x d yI would really appreciate any help!

Question

Given the step function  g  (  x  ,  y  )  =      {                                        1                          x              2                                                0          &amp;lt;          y          &amp;lt;          x          &amp;lt;          1                                      −                      1                          y              2                                                0          &amp;lt;          x          &amp;lt;          y          &amp;lt;          1                                      0                          otherwise                        I want to show that      ∫          (      0      ,      1      )            ∫          (      0      ,      1      )        g  (  x  ,  y  )    d  λ  (  x  )    d  λ  (  y  )  ≠      ∫          (      0      ,      1      )            ∫          (      0      ,      1      )        g  (  x  ,  y  )    d  λ  (  y  )    d  λ  (  x  )However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as      ∫    0    x        ∫    0    1        1          x      2          d  x    d  y  −      ∫    0    1        ∫    0    y        1          y      2          d  x    d  yI also considered using limits such as following one, but again I end up with divergence.      lim          k      →      ∞            ∫          (              1        k            ,      1      )            lim          k      →      ∞            ∫          (              1        k            ,      1      )        g  (  x  ,  y  )    d  x    d  yI would really appreciate any help!

Josie Stephenson · Accepted Answer

∫                      (            0            ,            1            )                                    (                      ∫                          (              0              ,              1              )                                g          (          x          ,          y          )                    d          λ          (          x          )          )                        d        λ        (        y        )                            =                                                      ∫                      (            0            ,            1            )                                    (                      ∫                          (              0              ,              y              )                                −                      1                          y              2                                          d          λ          (          x          )          +                      ∫                          (              y              ,              1              )                                            1                          x              2                                          d          λ          (          x          )          )                        d        λ        (        y        )                            =                                                      ∫                      (            0            ,            1            )                                    (          −                      1            y                    +                                    1              −              y                        y                    )                        d        λ        (        y        )        =                  ∫                      (            0            ,            1            )                          −        1                d        λ        (        y        )        =        −        1.

Given the step function g ( x , y ) = { <mtable columnalign="l

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