Does anyone know how to approach this?Let f : R ↦ R be a measurable periodic function with period T &gt; 0. Show that if f ∈ L 1 ( [ 0 , T ] ) ,, then lim x → ∞ 1 x ∫ [ 0 , x ] f ( t ) d m ( t )exists. I am trying to prove this and have been getting nowhere. Since we are letting x get large, I am thinking we can split the integral up into intervals of its period. However, just because f is integrable on [ 0 , T ] does not mean that the integral on its whole domain will be finite.

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Does anyone know how to approach this?Let   f  :      R    ↦      R   be a measurable periodic function with period   T  &amp;gt;  0. Show that if   f  ∈      L    1    (  [  0  ,  T  ]  )  ,, then      lim          x      →      ∞            1    x        ∫          [      0      ,      x      ]                f  (  t  )     d  m  (  t  )exists. I am trying to prove this and have been getting nowhere. Since we are letting   x get large, I am thinking we can split the integral up into intervals of its period. However, just because   f is integrable on   [  0  ,  T  ] does not mean that the integral on its whole domain will be finite.

dariajoq9 · Accepted Answer

Let   x  &amp;gt;  0 be arbitrary. Write   x  =  n  T  +  r where   r  =  x  mod  T  ∈  [  0  ,  T  ). We have                              1          x                          ∫                      [            0            ,            x            ]                          f        (        t        )                d        m        (        t        )                            =                  1                      n            T            +            r                                    (          n                      ∫                          [              0              ,              T              ]                                f          (          t          )                    d          m          (          t          )          +                      ∫                          [              0              ,              r              ]                                f          (          t          )                    d          m          (          t          )          )                                                  =                  n                      n            T            +            r                                    ∫                      [            0            ,            T            ]                          f        (        t        )                d        m        (        t        )        +                  1                      n            T            +            r                                    ∫                      [            0            ,            r            ]                          f        (        t        )                d        m        (        t        )        .            I leave to you to show that       n          n      T      +      r        →      1    T   and       |        1          n      T      +      r            ∫          [      0      ,      r      ]        f  (  t  )    d  m  (  t  )      |    →  0 as   x  →  ∞. Hence      lim          x      →      ∞            1    x        ∫          [      0      ,      x      ]        f  (  t  )    d  m  (  t  )  =      1    T        ∫          [      0      ,      T      ]        f  (  t  )    d  m  (  t  )  .

Jorge Lawson · Accepted Answer

I claim that infact      lim          x      →      ∞            1    x        ∫    0    x    f  (  t  )  d  t  =      1    T        ∫    0    T    f  (  t  )  d  t  .Because   f is periodic, you know that for   i  ∈      N  ,      ∫          T      i              T      (      i      +      1      )        f  (  t  )  d  t  =      ∫    0    T    f  (  t  )  d  t  .In particular, for any   x  ∈  (  T  k  ,  T  (  k  +  1  )  ), we have that      ∫    0    x    f  (  t  )  d  t  =      ∑          i      =      0              k      −      1            ∫          T      i              T      (      i      +      1      )        f  (  t  )  d  t  +      ∫          T      k              x        f  (  t  )  d  t  =  k      ∫    0    T    f  (  t  )  d  t  +      ∫          0              x      −      T      k        f  (  t  )  d  t  .Thus we have that      |        1    x        ∫    0    x    f  (  t  )  d  t  −      1    T        ∫    0    T    f  (  t  )  d  t      |    ≤  (      1    T    −      k    x    )      |        ∫    0    T    f  (  t  )  d  t      |    +      1    x        |        ∫    0          x      −      T      k        f  (  t  )  d  t      |    ≤  (      1    T    −      k    x    )      |        ∫    0    T    f  (  t  )  d  t      |    +      1    x        ∫    0          T            |    f  (  t  )      |    d  t  .As   x  →  ∞, we see that       lim          x      →      ∞            k    x    =      lim          x      →      ∞                  [      x              /            T      ]        x    =      1    T   where [  y] denotes the floor of   y. From this we see that      lim          x      →      ∞            |        1    x        ∫    0    x    f  (  t  )  d  t  −      1    T        ∫    0    T    f  (  t  )  d  t      |    =  0  ,which establishes the desired claim.

Does anyone know how to approach this? Let f : <mrow class="MJX-TeXAtom-ORD"> <mi ma

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