# How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem?

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

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.