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Let $f \in L^1(-\infty, \infty)$. How to find the limit: $$ \lim_{n \to \infty} \int_{-\infty}^{\infty} \frac{f(x)e^{nx} dx}{1+e^{nx}}. $$ What are the difference between the integration in Riemann integral and Lebesgue integral when we compute integrations? Thank you very much.

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2 Answers

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Hint 1: Apply dominated convergence theorem to compute this integral. Check that all necessary conditions are satisfied.

Hint 2: Prove the following equality $$ \lim\limits_{n\to\infty}\frac{f(x)e^{nx}}{1+e^{nx}}= \begin{cases} f(x)&\quad\text{ if }\quad x>0\\ 0.5 f(0)&\quad\text{ if }\quad x=0\\ 0&\quad\text{ if }\quad x<0\\ \end{cases} $$

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For a function $f$ put $f+ = f\vee 0$ and $f^- = (-f)\vee 0$. Then $f = f^+ - f^-$. If $\int f^+$ and $\int f^-$ are finite and $f$ is a.e. continuous, the Riemann and Lebesgue integrals agree. In particular, if $f$ is absolutely integrable in the Riemann sense, it agrees with the Lebesgue integral.

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Small nit: Riemann integrability requires (local) boundedness of the integrand. –  Harald Hanche-Olsen Aug 1 '12 at 18:35
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