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I'm self-learning Real Analysis using Real Analysis of Folland, and I got stuck on this problem.

Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X) < \infty$, and let $(X, \overline{\mathcal{M}}, \overline{\mu})$ be its completion. Suppose $f: X \rightarrow \mathbb{R}$ is bounded. Prove that $f$ is $\overline{\mathcal{M}}$-measurable iff there exist sequences $\{\phi_n\}$ and $\{\omega_n\}$ of $\mathcal{M}$-measurable simple functions such that $\phi_n\ \le f \le \omega_n$ and $\int(\omega_n - \phi_n)d \mu < n^{-1}$. In this case, $\lim{\int \phi_n d\mu} = \lim{\int \omega_n d\mu} = \int{fd\overline{\mu}}$

I thought about those functions which uses in Rienmann integral, but this problem only relates to Lebesgue integral, so I don't think they can help. Can anyone help me or give me some clue so I can solve this problem? Thanks so much your help. I really appreciate.

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    $\begingroup$ Can someone explain for me why I got downvote for this topic? $\endgroup$ – le duc quang Jul 19 '15 at 7:29
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Since the solution to this appears to be nowhere on the internet - a bit shocking given the ubiquity of the textbook and the (fairly) elementary nature of the problem in the context of Real Analysis - I'll post the whole thing. Hopefully no intractable errors.

Suppose $f$ is $\bar{\mathcal{M}}$-measurable and non-negative. Then there is an $\mathcal{M}$-measurable $g$ such that $f = g$ up to an $\mathcal{M}$-null set $N$. If $s = \sum_1^k a_j\cdot \chi_{A_j}$ is an arbitrary simple function, then $s^- = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + 0\cdot \chi_N$ and $s^+ = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + M \cdot \chi_N$ are simple functions, where $M$ is a bounding constant for $f(x)$.

As in Theorem $\textbf{2.10}$, write simple functions $$\phi_n = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}, \hspace{1cm} \psi_n = \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}$$ $$E_n^k = g^{-1}((k2^{-n}, (k+1)2^{-n}]), \hspace{1cm} F_n = g^{-1}((2^n, M))$$ Then these are simple functions converging pointwise to $g$ from above and below, respectively, in monotone fashion. Note that for $n$ large enough, $2^n > M$ so that for $n$ sufficiently large we can simply write $$\phi_n^- = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k \setminus N}, \hspace{1cm} \psi_n^+ = \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_N$$ Since $f = g$ away from $N$, $\phi_n^- \leq f \leq \psi_n^+$ for all $n$, and on $N$ itself, $\phi_n^- \equiv 0 \leq f \leq M \equiv \psi_n^+$.

It remains to be shown that $\int(\psi_n^+ - \phi_n^-) d\mu \rightarrow 0$. For $n$ sufficiently large, $$\int_X(\psi_n^+ - \phi_n^-) d\mu = \int_X(\psi_n^+ - \phi_n^-) d\bar{\mu} = \int_N(\psi_n^+ - \phi_n^-) d\bar{\mu} + \int_{N^c}(\psi_n^+ - \phi_n^-) d\bar{\mu} =$$ $$=\int_{N^c}(\psi_n^+ - \phi_n^-) d\bar{\mu} \leq \int_{N^c}|\psi_n^+ - f|+ |f - \phi_n^-| d\bar{\mu} \leq$$ $$ \leq \int_{N^c}|\frac{1}{2^n}|+ |\frac{1}{2^n}| d\bar{\mu} = \frac{1}{2^{n-1}} \cdot \bar{\mu}(N^c) = \frac{\bar{\mu}(X)}{2^{n-1}} = \frac{\mu(X)}{2^{n-1}} \rightarrow 0$$ Thus after reindexing $n$ if need be, the claim is proven for non-negative $f$. Taking positive and negative parts gives the general result.

Conversely, suppose such $\phi_n, \psi_n$ exist, and assume $f$ is non-negative. If $\psi_n = \sum_1^k a_i \cdot \chi_{A_i}$ with $a_i > M$ then we may simply adjust to the simple $\tilde{\psi_n}$ whose coefficients are the same as those of $\psi_n$ other than changing any such $a_i > M$ to $M$. Since $f$ is bounded by $M$, it is still the case that $f \leq \tilde{\psi_n}$. Construct monotone sequences of simple functions $\Phi_n = \max\lbrace \phi_1, \phi_2, \dots, \phi_n \rbrace, \Psi_n = \min \lbrace \tilde{\psi_1}, \tilde{\psi_2}, \dots, \tilde{\psi_n} \rbrace$. Then $0 \leq \Phi_n \leq f \leq \Psi_n \leq M$. Further, $$|\Psi_n - \Phi_n| \leq |\psi_n - \phi_n| \implies \int(\Psi_n -\Phi_n) d\mu \leq \int (\psi_n - \phi_n) d\mu \rightarrow 0$$ Since $\Phi_n, \Psi_n$ are bounded above by $M$ they have finite integrals and are thus in $L^1(\mu)$. In particular so are their limits, which are $\mathcal{M}$-measurable, so that $\Phi = \lim \Phi_n, \Psi = \lim \Psi_n$ are contained in $L^1(\mu)$. For all $n$, $$\int (\Psi - \Phi) d\mu \leq \int (\Psi_n - \Phi_n) d\mu \rightarrow 0$$ Thus $\Psi - \Phi = 0$ $\mu$-a.e., or said another way $\Psi$ and $\Phi$ differ only on an $\mathcal{M}$-null set. Since $\Phi \leq f \leq \Psi$, it follows that $f$ differs from either only on an $\mathcal{M}$-null set and so is $\bar{\mathcal{M}}$-measurable, since each is $\mathcal{M}$-measurable. Taking positive and negative parts gives the general result.

For the comment at the end of the exercise, the first equality was shown directly in each direction. For the second equality, taking $g \equiv M$ in the DCT for $\bar{\mu}$ it follows that since $\phi_n \rightarrow f$ $\bar{\mu}$-a.e. and $\phi_n \in L^1(\mu) \subset L^1(\bar{\mu})$ that $\int \phi_n d\mu = \int \phi_n d\bar{\mu} \rightarrow \int f d\bar{\mu}$.

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