$int_{0}^{infty}frac{sin(x)}{x}dx$ exists, but $int_{mathbb{R^+}}frac{sin(x)}{x}dlambda$ doesn't exist?
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$int_{0}^{infty}frac{sin(x)}{x}dx$ is riemann integrable. But how to prove that it's not Lebesgue integrable ? (I tried contradiction). If $f(x)=frac{sin(x)}{x}$ , I supppose that $fin L^1(mathbb{R^+})$ . then, I' ll define the function sequence : $f_n(x)=frac{sin( x)}{x}mathbb{1}_{[0,n]}(x).$ We have $f_n$ converges $lambda$ -a.e to $f$ . and $|f_n|leq|f|$ , for all $1leq n$ . and $f$ integrable (hypothesis). then with the dominated convergence theorem $$lim_nint_{0}^{n}f_ndlambda=int_{mathbb{R^+}}fdlambda$$ which is equivalent to $$int_{0}^{infty}frac{sin(x)}{x}dx=int_{mathbb{R^+}}f dlambda$$ Thats basically what I did.
integration lebesgue-integral
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