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Thread: Lebesgue integrable

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    Lebesgue integrable

    I'm trying to prove $\displaystyle f(x) = e^{-x^2} \in L^1(\mathbb{R})$ ; that is, that $\displaystyle f$ is Lebesgue integrable over $\displaystyle \mathbb{R}$.

    Let $\displaystyle f(x) = e^{-x^2} \chi_{(-\infty , \infty)}(x)$ and $\displaystyle f_n = f \chi_{[-n,n]}$.

    Since $\displaystyle f\geqslant 0$, $\displaystyle (f_n)$ is an increasing sequence of functions which converges everywhere to $\displaystyle f$.

    We want to show that $\displaystyle f$ is integrable so it suffices to show $\displaystyle \int f_n$ converges as $\displaystyle n\to\infty$. It then follows from the Monotone Convergence Theorem that $\displaystyle f\in L^1 (\mathbb{R})$ (and $\displaystyle \int f = \lim_{n\to\infty} \int f_n$).

    We use the given hint ("don't try to find $\displaystyle \int f$ ; integrate a simpler upper bound instead").

    We see that $\displaystyle f_n$ is bounded by ___. The integral of this upper bound converges, so by comparison test $\displaystyle \int f_n$ converges.

    My hurdle is filling the gap for an upper bound. $\displaystyle f_n$ is bounded by the constant function 1 but $\displaystyle \int_{-n}^n 1 dx= 2n$ which doesn't converge as $\displaystyle n \to\infty$. Any suggestions?
    Last edited by ProofbyInduction; Apr 16th 2012 at 07:26 AM.
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