An equivalent condition of $f^+ \in L^1 (\Omega)$ for $f \in L^1_{\text{loc}} (\Omega)$

Question

Let $\Omega$ be an open subset of $\mathbb R^d$ and $f \in L^1_{\text{loc}} (\Omega)$. I'm trying to do below exercise, i.e.,

Prove that $f^+ := \max\{f, 0\} \in L^1 (\Omega)$ IFF $$ B := \sup \left \{\int f \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \text{ and } \varphi \ge 0 \right\} < \infty. $$ If $f^+ \in L^1 (\Omega)$ then $B = \|f^+\|_1$.

1. Could you have a check on my below attempt?

2. Is there a shorter alternative approach?

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Let $$ A := \sup \left \{\int f^+ \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\}. $$

Let's show that $A=B$. Let $f^- := \max\{-f, 0\}$. Then $$ \begin{align} B &= \sup \left \{\int f^+ \varphi - \int f^- \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \text{ and } \varphi \ge 0 \right\} \\ &\le \sup \left \{\int f^+ \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \text{ and } \varphi \ge 0 \right\} \\ &\le \sup \left \{\int f^+ \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\} \\ &=A. \end{align} $$

Let's prove the reverse, i.e., $A \le B$. We need the following approximation result, i.e.,

Lemma 1 Let $\Omega$ be an open subset of $\mathbb R^d$. Let $u \in L^\infty (\Omega)$.

1. There is $(u_n) \subset \mathcal C_c^{\infty} (\Omega)$ such that

• $\sup_n \|u_n\|_\infty \le \|u\|_\infty$,
• $u_n \xrightarrow{n\to\infty} u$ a.e. on $\Omega$, and
• $u_n \xrightarrow{n\to\infty} u$ in $L^\infty (\Omega)$ w.r.t. the weak$^*$ topology $\sigma(L^\infty, L^1)$.

2. If $u \ge 0$ a.e. on $\Omega$, we can further take $u_n \ge 0$ a.e. on $\Omega$.
3. Deduce that $\mathcal C_c^{\infty} (\Omega)$ is dense in $L^\infty (\Omega)$ w.r.t. $\sigma(L^\infty, L^1)$.

Fix $\varphi \in \mathcal C^\infty_c(\Omega)$ such that $\|\varphi\|_\infty \le 1$. Let $P := \{x \in \Omega : \varphi (x) >0\}$. Then $P$ is open in $\Omega$. Fix $\varepsilon >0$. By Lemma 1, there is $\psi \in \mathcal C_c^{\infty} (P)$ such that $\psi \ge 0, \|\psi\|_\infty \le 1$ and $|\int_P f^+ \varphi - \int_P f \psi| < \varepsilon$. We define $\overline \psi:\Omega \to \mathbb R$ by $\overline \psi (x) := \psi (x)$ if $x \in P$ and $0$ otherwise. Then $\overline \psi \in \mathcal C_c^{\infty} (\Omega)$ such that $\overline \psi \ge 0, \|\overline \psi\|_\infty \le 1$ and $$ \int f^+ \varphi \le \int_P f^+ \varphi \le \varepsilon +\int_P f^+ \psi = \varepsilon +\int f^+ \overline \psi. $$

Let $\varepsilon \downarrow 0$, we get $$ A = \sup \left \{\int f^+ \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \text{ and } \varphi \ge 0 \right\}. $$

Fix $\varphi \in \mathcal C^\infty_c(\Omega)$ such that $\|\varphi\|_\infty \le 1$ and $\varphi \ge 0$. Let $P := \{x \in \Omega : f (x) >0\}$. Then $\int f^+ \varphi = \int_P f \varphi$. Fix $\varepsilon >0$. Because Lebesgue measure is outer regular, there is an open subset $Q$ of $\Omega$ such that $P \subset Q$ and $$ \bigg | \int_P f \varphi - \int_Q f \varphi \bigg |< \varepsilon. $$

By Lemma 1, there is $\psi \in \mathcal C_c^{\infty} (Q)$ such that $\psi \ge 0, \|\psi\|_\infty \le 1$ and $|\int_Q f \varphi - \int_Q f \psi| < \varepsilon$. We define $\overline \psi:\Omega \to \mathbb R$ by $\overline \psi (x) := \psi (x)$ if $x \in Q$ and $0$ otherwise. Then $\overline \psi \in \mathcal C_c^{\infty} (\Omega)$ such that $\overline \psi \ge 0, \|\overline \psi\|_\infty \le 1$ and $$ \int f^+ \varphi \le \varepsilon +\int_Q f \varphi \le 2 \varepsilon + \int_Q f \psi = 2\varepsilon +\int f \overline \psi. $$

Let $\varepsilon \downarrow 0$, we get $$ \begin{align} A &\le \sup \left \{\int f \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \text{ and } \varphi \ge 0 \right\} \\ &= B. \end{align} $$

The claim then follows from below result, i.e.,

Lemma 2 $f \in L^1 (\Omega)$ IFF $$ A := \sup \left \{\int f \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\} < \infty. $$ If $f \in L^1 (\Omega)$ then $A = \|f\|_1$.

Answer 1

There is another interesting result along these lines:

Suppose $X$ is a locally compact Hausdorff space (l.c.H), and $\mu$ a Radon measure defined on the Borel sets $\mathscr{B}(X)$ in $X$. Denote by $C_{00}(X)$ the space of all real valued continuous functions on $X$ that have compact support, and by $C_{00}(X;\mathbb{C})$ the space of all complex-valued continuous functions of compact support.

Theorem: Let $f\in L^{loc}_1(\mu)$ and $g\in C^+_{00}(X)$. If $f$ is real valued, then \begin{align} \int f^+g\,d\mu&=\sup\left\{\int fh\,d\mu: h\in C_{00}(X),\, 0\leq h\leq g\right\} \tag{1}\label{one}\\ \int |f|g\,d\mu&=\sup\left\{\int fh\,d\mu: h\in C_{00}(X),\, |h|\leq g\right\} \tag{2}\label{two} \end{align} If $f$ is complex, then \begin{align} \int |f|g\,d\mu&=\sup\left\{\Big|\int fh\,d\mu\big|: h\in C_{00}(X;\mathbb{C}),\, |h|\leq g\right\}\tag{3}\label{three} \end{align} The space $C_{00}(X)$ can be replace by other space of bounded functions that is dense in $L_1(\mu)$; for example, if $X=[0,1]$, one my consider step functions.

Proof: Without loss of generality, we may assume that $f\in L_1(X)$. For $\varepsilon>0$, there is $f_1\in C_{00}(X)$ such that $$\int|f-f_1|g\,d\mu<\frac{\varepsilon}{3}$$ Notice that if $f$ is real, then $f_1$ can be chosen to be real, which we will do if that case. Let $K=\operatorname{supp}(f_1)$ and let $U$ be an open set with compact closure such that $K\subset U$. Let $h\in C_{00}(X)$ such that $0\leq h\leq 1$, $h=1$ on $K$ and $h=0$ on $X\setminus U$. For $\delta>0$, define the function $$\phi_\delta=\frac{\overline{f_1}}{|f_1|+\delta h}$$ on the support of $h$ and $0$ otherwise; and if $f$ is real, $$\psi_\delta=\frac{f^+_1}{f^+_1 +\delta h}$$ on the support of $h$ and $0$ otherwise. Since $\phi_\delta=0=\psi_\delta$ on $U\setminus K$, $\phi_\delta$ and $\psi_\delta$ are continuous and of compact support; by definition, $|\phi_\delta|\leq1$ and $0\leq \psi_\delta\leq 1$. Notice that $\phi_\delta\xrightarrow{\delta\rightarrow0}\frac{\overline{f_1}}{|f_1|}$ and $\psi_\delta\xrightarrow{\delta\rightarrow0}\mathbb{1}_{\{f_1>0\}}=\frac{f^+_1}{f_1}$ pointwise on $X$.

If $f$ is real, then by dominated convergence, there is $\delta>0$ small enough so that $$ \int|f^+_1g-f_1\psi_\delta g|<\varepsilon/3 $$ Then \begin{align} \int f^+g-f\psi_\delta g&\leq \int (f^+-f^+_1)g+\int f^+_1g-f_1\psi_\delta g+ \int f_1\psi_\delta g-f\psi_\delta g\\ &\leq2\int|f-f_1|g+\int|f^+_1g-f_1\psi_\delta g|<\varepsilon \end{align} Hence $$\int f^+g\,d\mu<\int f\psi_\delta g\,d\mu+\varepsilon$$ This implies \eqref{one}.

Generally, for $f$ complex, dominated convergence yields $\delta>0$ small enough so that \begin{align} \int \big||f_1|g -f_1 \phi_\delta g\,d\mu\big|\,d\mu<\varepsilon/3 \end{align} Then \begin{align} \Big|\int(|f|g- f\phi_\delta g)\Big| &=\left|\int(|f|-|f_1|)g + \int|f_1|g- f_1\phi_\delta g +\int f_1\phi_\delta g-f\phi_\delta g\right|\\ &\leq\int|f-f_1|g +\int\big||f_1|g -f_1 \phi_\delta g\big|+\int|\phi_\delta||f-f_1|g\\ &\leq2\Big(\int|f-f_1|g\Big)+\frac{\varepsilon}{3}<\varepsilon \end{align} Hence $$\int|f|g\,d\mu\leq \big|\int f\phi_\delta g\,d\mu\big|+\varepsilon$$ This implies that \eqref{three} (if $f$ is real, \eqref{two} follows in a similar fashion).

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Comment: The problem in the OP can be derived from Theorem 1 above by first approximating as follows. Choose $f_1\in C_{00}(X)$ such that $\|f-f\_1\|_1<\varepsilon/2$. Let $K=\operatorname{supp}(f_1)$, and $h\in C_{00}(X)$ with $0\leq h\leq 1$ such that $h=1$ on $K$ and $0$ outside a compact neighborhood $U$ of $K$, and let $f_2=f_1h$. Notice that $f_1=f_1=0$ and so \begin{align} \int|f-f_1h|<\varepsilon/2 \end{align} Apply Theorem 1 to $f_1h$.

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