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Thread: Two Lebesgue Integration Problems

  1. #1
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    Two Lebesgue Integration Problems

    1. Let f be a nonnegative measurable function on $\displaystyle \Re$. Show that if $\displaystyle F : \Re \rightarrow \Re$ is defined by $\displaystyle \int_{(0,x]} f(t)dm(t)$ then F is continuous on $\displaystyle \Re$.

    I know that to prove that a function is continuous, you have to show that for any open set $\displaystyle A \subseteq \Re$ so that $\displaystyle f^{-1}(A)$ is open, but I'm not sure what to do from there.

    2. Let f be a nonnegative measurable function on a measurable set E with $\displaystyle \int_E f(x)dm(x)<\infty$. Prove that for each $\displaystyle \epsilon>0$ there exists $\displaystyle \delta>0 $ such that for every measurable set $\displaystyle A \subseteq E$ with $\displaystyle m(A) < \delta$ we have $\displaystyle \int_A f(x)dm(x) < \epsilon$

    I am clueless how to even start this problem.
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  2. #2
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    For 1. use the sequential characterization of continuity. Take a sequence {x_n} converging to x. It suffices to show F(x_n) --> F(x).
    A hint to do this is bring (0,x_n] in as a characteristic function and use an appropriate convergence theorem.
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  3. #3
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    Quote Originally Posted by grad444 View Post
    2. Let f be a nonnegative measurable function on a measurable set E with $\displaystyle \int_E f(x)dm(x)<\infty$. Prove that for each $\displaystyle \epsilon>0$ there exists $\displaystyle \delta>0 $ such that for every measurable set $\displaystyle A \subseteq E$ with $\displaystyle m(A) < \delta$ we have $\displaystyle \int_A f(x)dm(x) < \epsilon$

    I am clueless how to even start this problem.
    For 2.: Let $\displaystyle \varepsilon>0$. Using the bounded convergence theorem, prove that there exists $\displaystyle M$ such that $\displaystyle \int_{\{f>M\}} f(x)dm(x)<\varepsilon/2$ (integration on the set where $\displaystyle f(x)>M$). Then choose $\displaystyle \delta<\varepsilon/(2M)$, and conclude using:

    $\displaystyle \int_A f(x)dm(x)= \int_{A\cap\{f>M\}} f(x)dm(x)+\int_{A\cap\{f\leq M\}} f(x)dm(x)\leq \varepsilon/2 + M m(A)$.
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