Let I = [ a , b ] with a &lt; b and let u : I → R be a function with bounded pointwise variation, i.e. V a r I u = sup { ∑ i = 1 n | u ( x i ) − u ( x i − 1 ) | } &lt; ∞where the supremum is taken over all partition P = { a = x 0 &lt; x 1 &lt; . . . &lt; x n − 1 &lt; b = x n }. How can I prove that if u satisfies the intermediate value theorem (IVT), then u is continuous?My try: u can be written as a difference of two increasing functions f 1 , f 2 . I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that f 1 , f 2 satisfies the (ITV) the assertion follows. But, is this true? I mean, f 1 , f 2 satisfies (ITV)?

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Let   I  =  [  a  ,  b  ] with   a  &amp;lt;  b and let   u  :  I  →      R   be a function with bounded pointwise variation, i.e.  V  a      r    I    u  =  sup  {      ∑          i      =      1        n        |    u  (      x    i    )  −  u  (      x          i      −      1        )      |    }  &amp;lt;  ∞where the supremum is taken over all partition   P  =  {  a  =      x    0    &amp;lt;      x    1    &amp;lt;  .  .  .  &amp;lt;      x          n      −      1        &amp;lt;  b  =      x    n    }. How can I prove that if   u satisfies the intermediate value theorem (IVT), then   u is continuous?My try:   u can be written as a difference of two increasing functions       f    1    ,      f    2  . I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that       f    1    ,      f    2   satisfies the (ITV) the assertion follows. But, is this true? I mean,       f    1    ,      f    2   satisfies (ITV)?

Waylon Padilla · Accepted Answer

Begin by proving that every function of bounded variation has finite one-sided limits at every point. (Decomposition into monotone functions does this in one line.) Then observe that the intermediate value property fails unless   f  (  a  +  )  =  f  (  a  −  )  =  f  (  a  ).

Let I = <mo stretchy="false">[ a , b <mo stretchy="false">] with a &l

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