I'm trying to solve the following problem.Let f be an integrable function in (0,1). Suppose that ∫ 0 1 f g ≥ 0for any non negative, continuous g : ( 0 , 1 ) → R . Prove that f ≥ 0 a.e. in (0,1).I'm a little unsure on what it is that I must prove in order to conclude that f ≥ 0. I tried to show that ∫ 0 1 f 2 ≥ 0 but I couldn't get very far.I'm seeking hints on how to solve this. Thanks.

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I&#039;m trying to solve the following problem.Let   f be an integrable function in (0,1). Suppose that      ∫    0    1    f  g  ≥  0for any non negative, continuous   g  :  (  0  ,  1  )  →      R  . Prove that   f  ≥  0 a.e. in (0,1).I&#039;m a little unsure on what it is that I must prove in order to conclude that   f  ≥  0. I tried to show that       ∫    0    1        f    2    ≥  0 but I couldn&#039;t get very far.I&#039;m seeking hints on how to solve this. Thanks.

Nirdaciw3 · Accepted Answer

Suppose that   A  ⊂  (  0  ,  1  ) is measurable, is of positive measure and   f  &amp;lt;  0 on   A. The idea is that we want to construct a continuous function   g such that      ∫    0    1    f  g    d  x  &amp;lt;  0.A logical way to do this would be to choose   g such that   g  ⩾  0 in   A and   g  =  0 on   (  0  ,  1  )  ∖  A. However, since   A is only a measurable set, in general   g will be discontinuous.There exists a (relatively) closed set   F  ⊂  A such that   |  A  ∖  F  |  &amp;lt;  ϵ. Choosing   ϵ  =  |  A  |      /    2  &amp;gt;  0, we have that  |  F  |  =  |  A  |  −  |  A  ∖  F  |  =  |  A  |      /    2  &amp;gt;  0.Since   |  F  |  &amp;gt;  0, the interior of   F is nonempty. Thus, there exists an open set   U compactly contained in the interior of   F (just take a small ball for example). Define   g such that   g is continuous, nonnegative,   g  =  0 in   A  ∖  F, and   g  =  1 in   U. Then                              ∫          0          1                f        g                d        x                            =                  ∫          F                f        g                d        x                                          =                  ∫          U                f                d        x        +                  ∫                      F            ∖            U                          f        g                d        x                                          ⩽                  ∫          U                f                d        x                                          &amp;lt;        0.            This completes the proof.

I'm trying to solve the following problem. Let f be an integrable function in (0,1). Suppose th

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