How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:20:14Z http://mathoverflow.net/feeds/question/10374 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10374/how-can-we-use-the-bounded-convergence-theorem-in-this-proof-of-the-riesz-represe How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem? S. Donovan 2010-01-01T14:46:16Z 2010-01-01T15:05:22Z <p>I'm studying the proof of the <strong>Riesz Representation Theorem</strong> as it appears in Ch. 6 of Royden's <em>Real Analysis</em>. When I looked on the web I noted there are a few different theorems that go by the name "Riesz Representation Theorem" so I'll state the one I'm looking at:</p> <blockquote> <p>Let F be a bounded linear functional on $L^p$, $1 \leq p &lt; \infty$. Then there is a function $g \in L^q \ni$, $F(f) = \int fg$.</p> </blockquote> <p>The proof starts by showing that g exists for the characteristic functions $\chi_s = \chi_{[0,s]}$. Then we can write any step function as the sum of $\chi_{s_{i}}$. So based on an earlier theorem, if f is a bounded measurable function in [0,1] we can write a sequence of step functions, $&lt;\psi_n>$ that converge to f almost everywhere. This is the sentence that confuses me:</p> <blockquote> <p>"Since the sequence $&lt;|f- \psi_n|^p>$ is uniformly bounded and tends to zero almost everywhere, the bounded convergence theorem implies that $||f-\psi_n||_p \rightarrow 0$."</p> </blockquote> <p>But, when I look at the <strong>bounded convergence theorem</strong>, it would require $\mathop{\lim}\limits_{n \to \infty} |f-\psi_n|^p = 0$. Period. Not just almost everywhere, to get $\mathop{\lim}\limits_{n \to \infty} \int_{[0,1]}|f-\psi_n|^p = \int_{[0,1]} 0 = 0$.</p> <p>So, that's where I'm stuck. I just don't see how the bounded convergence theorem can work here. (Side question: I also, don't feel I really know what Royden means by "uniformly bounded" is that just saying there is one bound that works for the whole set? How is that different from regular bounded?)</p> http://mathoverflow.net/questions/10374/how-can-we-use-the-bounded-convergence-theorem-in-this-proof-of-the-riesz-represe/10376#10376 Answer by fedja for How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem? fedja 2010-01-01T15:05:22Z 2010-01-01T15:05:22Z <p>Such questions should be really asked on AoPS rather than here, but, once you've already posted it on MO, I'll answer.</p> <p>1) The set of zero measure can always be ignored when performing Lebesgue integration, so to say <code>$g_n\to 0$</code> everywhere or almost everywhere is practically the same: just drop the measure zero set where the convergence fails and apply the bounded convergence theorem as you know it to the integral over the rest.</p> <p>2) Yes, "uniformly bounded" means here that there is one bound for all functions simultaneously. In this context there is any difference between saying "uniformly bounded sequence" and "bounded sequence" but there is a clear difference between saying "a sequence of bounded functions" and "a sequence of uniformly bounded functions".</p>