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Theorem 5 Let $[a,b]$ be a closed, bounded interval and $1 \le p < \infty$. Suppose that $T$ is a bounded linear functional on $L^p[a,b]$. Then there is a function $g \in L^q[a,b]$, where $q$ is the conjugate of $p$, for which $$T(f) = \int_a^b g \cdot f \mbox{ for all } f \in L^p[a,b]$$

Here is the part I want to focus on:

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At this point, the author has already shown that there exists an integral function $g$ over $[a,b]$ for which $T(f) = \int_{a}^{b} g \cdot f$ for all step functions $f$ on $[a,b]$.

My first question is, why can the sequence $\{\varphi_n\}$ be taken as uniformly pointwise bounded on $[a,b]$? I looked at the proof of proposition 10 and discovered uniform boundedness mentioned nowhere. Second, how can we conclude that $$\lim_{n \to \infty} \int_a^b g \cdot \varphi_n = \int_a^b g \cdot f$$ from the Dominated Convergence Theorem if it isn't known that $g \cdot \varphi_n \to g \cdot f$ pointwise a.e. on $[a,b]$? We only have that $\varphi_n \to f$ in $L^p[a,b]$, not pointwise. I imagine that $|g \cdot f |$ is the dominating function, but I'm not sure I see why.

My last question is, why bother doing any of this? If the linear functional $f \mapsto \int_a^b g \cdot f$ agrees with $T$ on the set of step functions, which is does, then they agree on $L^p[a,b]$, since the step functions are dense in $L^p[a,b]$. So what is the point of showing this further fact that they agree on the simple functions?

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  • $\begingroup$ Do you know where in the book step function is defined? $\endgroup$ – grndl May 3 '18 at 1:34
  • $\begingroup$ If $\{\phi_n\}$ is uniformly bounded by $B$, then $\{g \cdot \phi_n\}$ is bounded by $g \cdot B$. To apply dominated convergence, it suffices that $\phi_n \to f$ in measure, which is implied by convergence in $L^p$. $\endgroup$ – grndl May 3 '18 at 1:42
  • $\begingroup$ @aduh Are you referring to Royden too? I had trouble finding the definition also. In fact, I asked this question a while back because I couldn't find it. A step function is a simple function except the domains of the indicator functions are intervals, rather than general measurable sets. $\endgroup$ – user193319 May 3 '18 at 11:13
  • $\begingroup$ @aduh Is this a correct proof that convergence in $L^p$ implies convergence in measure? Let $\epsilon > 0$, and suppose $f_n \to f$ in $L^p$. Then $$m\{x \in E \mid |f_n(x) - f(x)| \ge \epsilon\} = m\{x \in E \mid |f_n(x) - f(x)|^p \ge \epsilon^p \} \le \frac{1}{\epsilon^p} \int_{E} |f_n-f|^p \to 0$$ as $n \to \infty$. $\endgroup$ – user193319 May 3 '18 at 11:29
  • $\begingroup$ @aduh If $\varphi_n \to f$ in measure, then there is a subsequence $\{\varphi_{n_k}\}$ converging pointwise to $f$. So, $\{\varphi_{n_k}\}$ is a sequence of bounded functions converging pointwise to the bounded function $f$. Unfortunately, I don't think this implies either $\{\varphi_{n_k}\}$ or $\{\varphi_{n}\}$ is uniformly bounded... $\endgroup$ – user193319 May 3 '18 at 14:58

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