So a well-known theorem from Lebesgue integration is the dominated convergence theorem. It talks about a sequence [itex]f_1,f_2,\ldots[/itex] of functions converging pointwise to a function [itex]f[/itex]. And if [itex]|f_n(x)| \leq g(x)[/itex] for an integrable function [itex]g[/itex], then we have [itex]\int f_n \to \int f[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

But what if we have a given function [itex]f_\epsilon[/itex] which depends on some parameter [itex]\epsilon[/itex], which we are taking to zero? Suppose you've shown that [itex]f_\epsilon \to 0[/itex] pointwise as [itex]\epsilon \to 0[/itex]; however, you know [itex]f_\epsilon[/itex] is NOT bounded in [itex]\epsilon[/itex] - if you take [itex]\epsilon[/itex] large enough, you can make [itex]\sup_{x\in \mathbb R} f_\epsilon(x)[/itex] arbitrarily large. But, of course, we don't WANT to do that - we want to take [itex]\epsilon[/itex] small. And suppose you know that [itex]\epsilon < \epsilon_0[/itex] means [itex]\sup f_{\epsilon} < \sup f_{\epsilon_0}[/itex]. My question is this: Can we still apply the DCT to show that [itex]\lim_{\epsilon \to 0} \int_{\mathbb R} f_\epsilon = 0[/itex]? If so, why?

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# Confusion regarding the dominated convergence theorem

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