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Math Help - Prove that this integral function is continuous.

  1. #1
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    Prove that this integral function is continuous.

    If  f \in L^1(m) and F(x) = \int _{- \infty}^{x}f(t)dt , then F is continuous on  \mathbb {R}

    Proof so far.
    Let x_0 \in X ,

    I need to show that  \forall \epsilon > 0 , there exists  \delta > 0 such that whenever  \mid x - x_0 \mid < \delta where  x \in X , then I will have  \mid F(x)-F(x_0) \mid < \epsilon

    Now, I know have that  \mid F(x)-F(x_0) \mid = \mid \int _{- \infty}^{x}f(t)dt - \int _{- \infty}^{x_0}f(t)dt \mid =  \mid \int _{x_0}^{x}f(t)dt \mid

    Since f is integrable, this integral exists, but how should I refine it so it is less than  \epsilon ?

    Thank you.
    Last edited by tttcomrader; November 15th 2009 at 06:02 AM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    [x_0,x] is compact, so f(t) has a max on that interval.

    Note: This only works in the reals. I notice you have X as your space, so I'm not sure this applies, unless you know that closed intervals are compact.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    [x_0,x] is compact, so f(t) has a max on that interval.
    How about f(t)= x^{-1/3} if 0<\vert x \vert \leq 1 and 0 otherwise. Then f \in L^1( \mathbb{R} ) but in any compact set of the form [-a,b] with a,b \geq 0 we have that f is not bounded.
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  4. #4
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    Sorry, I was thinking that f was continuous.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    If  \in L^1(m) and F(x) = \int _{- \infty}^{x}f(t)dt , then F is continuous on  \mathbb {R}

    Proof so far.
    Let x_0 \in X ,

    I need to show that  \forall \epsilon > 0 , there exists  \delta > 0 such that whenever  \mid x - x_0 \mid < \delta where  x \in X , then I will have  \mid F(x)-F(x_0) \mid < \epsilon

    Now, I know have that  \mid F(x)-F(x_0) \mid = \mid \int _{- \infty}^{x}f(t)dt - \int _{- \infty}^{x_0}f(t)dt \mid =  \mid \int _{x_0}^{x}f(t)dt \mid

    Since f is integrable, this integral exists, but how should I refine it so it is less than  \epsilon ?

    Thank you.
    Pro tip; \int_{x_0}^{x}f d\mu=\int_X f \mathbf{1}_{[x_0,x]} d\mu. What does Fatou say when you look at x\rightarrow x_0.
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    Fatou's Lemma says that  \int liminf  \ f_n \leq lim inf \int f_n

    So as  x_n \rightarrow x_0 ,  \int  f 1_{ [ x_0,x ] }  = \int lim inf \  f 1_{ [x_0,x_n]} \leq lim inf \int f 1_{ [x_0,x_n]} = 0
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  7. #7
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    Quote Originally Posted by tttcomrader View Post
    Fatou's Lemma says that  \int liminf  \ f_n \leq lim inf \int f_n

    So as  x_n \rightarrow x_0 ,  \int  f 1_{ [ x_0,x ] }  = \int lim inf \  f 1_{ [x_0,x_n]} \leq lim inf \int f 1_{ [x_0,x_n]} = 0
    Sorry I ment to say reverse Fatou.

    \lim\sup \int f\mathbf{1}_{[x_0,x_n]} d\mu \leq \int \lim \sup f\mathbf{1}_{[x_0,x_n]}d\mu=\mu(f \mathbf{1}_{x_0})=0
    for any sequence  x_n \rightarrow x.

    Now if you do apply what that means in epsilon and deltas, you get continuity.
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    I have never use the reverse Fatou's lemma. I wonder if this proof can work as well.

    Let  \{ x_n \} \rightarrow x_0 .

    Then  \lim _{n \rightarrow \infty } \mid F(x_n)-F(x_0) \mid = \lim _{n \rightarrow \infty } \mid \int ^{x_0}_{x_n} f(t)dt \mid = \lim _{n \rightarrow \infty } \int f 1_{[x_n,x_0]}d \mu = \int \lim _{n \rightarrow \infty } f 1_{[x_n,x_0]}d \mu = 0

    Will this be okay as well? Thank you.
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  9. #9
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    Quote Originally Posted by tttcomrader View Post
    I have never use the reverse Fatou's lemma. I wonder if this proof can work as well.

    Let  \{ x_n \} \rightarrow x_0 .

    Then  \lim _{n \rightarrow \infty } \mid F(x_n)-F(x_0) \mid = \lim _{n \rightarrow \infty } \mid \int ^{x_0}_{x_n} f(t)dt \mid = \lim _{n \rightarrow \infty } \int f 1_{[x_n,x_0]}d \mu = \int \lim _{n \rightarrow \infty } f 1_{[x_n,x_0]}d \mu = 0

    Will this be okay as well? Thank you.
    Well you can't really take the limit inside, you could however use monotone/dominated convergence. If you have not done those tell me what you have done so I can be more helpful.
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    Yes, I have done both Monotone and DCT.

    But since  1_{[x_n,x_0]} is approaching zero, isn't f 1_{[x_n,x_0]} \leq f for all n? Thus I can use the monotone convergence theorem?
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  11. #11
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    Quote Originally Posted by tttcomrader View Post
    Yes, I have done both Monotone and DCT.

    But since  1_{[x_n,x_0]} is approaching zero, isn't f 1_{[x_n,x_0]} \leq f for all n? Thus I can use the monotone convergence theorem?
    Be careful, \mathbf{1}_{[x_n,x_0]} \rightarrow \delta_{x_0} \neq 0 where \delta is the Dirac delta function (1 at x_0, 0 everywhere else). Now the Dirac delta function has Lebesgue measure 0 (i.e. integral 0). The sequence f\mathbf{1}_{[x_n,x_0]} is dominated by f (which is integrable), so DMC implies that the integral converges to 0, because DMC tells you that the integral of your sequence converges to \int f\delta_{x_0}=0.
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