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I have a question about the following question:

Let $f:\mathbb{R} \to \mathbb{R}$ be a measurable function. We know there exists a constant $K$ such that for every bounded continuous function $h:\mathbb{R} \to \mathbb{R}$ we have $fh$ intebrable and $$\int_\mathbb{R} f(x)g(x) dx \leq K ||g||_\infty$$ Show $f$ is integrable and $$ \int_\mathbb{R} |f(x)| dx \leq K$$

My initial though was to use sign of $f$ in some way, but I have know idea were to start. Could you guys help me? Thanks!

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I'm guessing you meant to have $g$ instead of $h$ in your problem statement. The integrability of $f$ is immediate from the hypothesis if you take $g = 1$. $g = 1$ is bounded and continuous and moreover $fg$ is integrable by assumption, meaning that $\int_{\Bbb R}|f(x)g(x)|\,dx < \infty$. However $|f(x)g(x)| = |f(x)|$ and so $f$ is integrable. The hard part is showing that $\int_{\Bbb R}|f(x)|\,dx \le K$.

You're definitely on the right track with considering the sign of $f$. $\operatorname{sgn}(f)$ need not be a continuous function so we cannot just let $g$ be $\operatorname{sgn}(f)$ and hope to apply the hypothesis in your post. $\operatorname{sgn}(f)$ is of course measurable.

Since we would like to be able to apply your hypothesis, we need to somehow approximate $\operatorname{sgn}(f)$ with continuous functions, but this cannot be guaranteed if we do not work on a finite interval$^{[1]}$. So let us simply consider $g = \chi_{[a,b]}\operatorname{sgn}(f)$. This function can be approximated almost everywhere with continuous functions since we've effectively restricted ourselves to a finite interval. (See Egoroff/Lusin.)

Let $g_n$ be a sequence of continuous functions on $[a,b]$ converging to $g$ almost everywhere. We can without loss of generality assume that $|g_n|\le 1 = |g|$ almost everywhere - meaning that we are approximating from within (this is similar to the usual approach with simple functions). Then $fg_n \to fg$ pointwise almost everywhere on $[a,b]$ and moreover $|fg_n| \le |fg|$ so we can use Lebesgue dominated convergence since $|fg|$ is integrable. That is

$$ \int_{\Bbb R} f(x)g(x)\,dx = \lim_n \int_{\Bbb R} f(x)g_n(x)\,dx.$$

I'll let you take it from here. There isn't much left to do at this point.


$^{[1]}$ This is not quite true, but the usual statements for Egoroff and Lusin restrict to finite intervals. If you are working in a sufficiently nice topological space, you can easily extend it to work on the whole space.

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  • $\begingroup$ But sign function is not continuous, which violates the requirement of $g$ being continuous and bounded, no? $\endgroup$
    – Ninanina
    Nov 11, 2015 at 21:22
  • $\begingroup$ Oh d'oh I somehow missed that part of your original post. There is a way to correct it but it'll take me about an hour to get to it. $\endgroup$ Nov 11, 2015 at 22:02
  • $\begingroup$ I have edited my answer to reflect the appropriate changes. Thanks for catching my mistake earlier. $\endgroup$ Nov 12, 2015 at 2:56

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