# Lebesgue integrable

• Apr 16th 2012, 07:23 AM
ProofbyInduction
Lebesgue integrable
I'm trying to prove $f(x) = e^{-x^2} \in L^1(\mathbb{R})$ ; that is, that $f$ is Lebesgue integrable over $\mathbb{R}$.

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

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

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

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

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

My hurdle is filling the gap for an upper bound. $f_n$ is bounded by the constant function 1 but $\int_{-n}^n 1 dx= 2n$ which doesn't converge as $n \to\infty$. Any suggestions?