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I'm studying the proof of the Riesz Representation Theorem as it appears in Ch. 6 of Royden's Real Analysis. 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:

Let F be a bounded linear functional on $L^p$, $1 \leq p < \infty$. Then there is a function $g \in L^q \ni$, $F(f) = \int fg$.

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, $<\psi_n>$ that converge to f almost everywhere. This is the sentence that confuses me:

"Since the sequence $<|f- \psi_n|^p>$ is uniformly bounded and tends to zero almost everywhere, the bounded convergence theorem implies that $||f-\psi_n||_p \rightarrow 0$."

But, when I look at the bounded convergence theorem, it would require $\mathop{\lim}\limits_{n \to \infty} |f-\psi_n|^p = 0$. Period. Not just almost everywhere, to get $\mathop{\lim}\limits_{n \to \infty} \int_{[0,1]}|f-\psi_n|^p = \int_{[0,1]} 0 = 0$.

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?)

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Try multiplying each element of your sequence by $\chi_S$, where $S$ is a measurable subset of [0,1] that has full measure and on which the sequence converges to zero. –  Yemon Choi Jan 1 '10 at 14:56
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1 Answer

up vote 2 down vote accepted

Such questions should be really asked on AoPS rather than here, but, once you've already posted it on MO, I'll answer.

1) The set of zero measure can always be ignored when performing Lebesgue integration, so to say $g_n\to 0$ 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.

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".

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Thanks. That really clears it up. I don't often get responses on AoPS are there any other sites? –  S. Donovan Jan 1 '10 at 15:15
    
Where are you posting there? College Playground is the right place for this level stuff and most questions get answered there (though, we cannot take care of everything and tend to ignore routine homeworks). If you tell me your AoPS screen name, I'll watch for your posts there. Note that AoPS is not a hotline, quite the opposite, so it is not unusual to get an answer there long after the question is asked. Also, it is not, strictly speaking, an "ask a question" site but rather a place to hang out and to have math. discussions. Still, it is the highest quality math. site for students I know of. –  fedja Jan 1 '10 at 15:38
    
Since you mention "College Playground" I see you are talking about the other aops site. I go by futurebird on the physics forum (which is more active than aops) and the "art of problem solving" livejournal (community.livejournal.com/mathematics ) which is very active, but has no latex and more of a place for fun math chat than questions. (I mean it's livejournal... though some good things do get done there..) The main AoPS site seemed well.. kind of dead. I'll give it another shot. Frankly, it's hard for me to tell what will be "too easy" and what won't. –  S. Donovan Jan 1 '10 at 16:22
    
Try looking up these proofs in other books first. If you can't find it anywhere else, then ask here. You can trust Anton more than the people who are actually here, since he's the administrator. If you think a question is really questionable, maybe go to meta and ask how you can improve it so it's fine for this site. The only problem I have with this question is that it seems like you didn't look it up in other books. –  Harry Gindi Jan 1 '10 at 18:27
    
Also, instead of AoPS, try #math on irc.efnet.net or #math on irc.freenode.net. these are IRC channels that tend to be fairly active and have a number of knowledgeable people. The only problem is that sometimes the channel is flooded with idiots asking how to "factor a polynomial" and other similar questions. –  Harry Gindi Jan 1 '10 at 18:36
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