For any bounded function $f:\mathbb{R}\to \mathbb{R}$, $$\int f=\inf \left\lbrace\int g : g\text{ is a simple function such that } f\le g \right\rbrace .$$

$f\le g$ implies that $\int f\le \inf \int g$.

Anyone can help me with finding a simple function $g\ge f$ with $\int g \le \int f + \epsilon$ ? Thank you.

  • $\begingroup$ Is this not immediate by definition? $\endgroup$ – bitesizebo Nov 1 '18 at 23:23
  • $\begingroup$ @bitesizebo for example in Zygmund's Measure and Integral, they define $\int_E f$ as the measure of the points under the graph of $f$ defined as $R(f,E) = \{(x,y) \in E \times \mathbb{R} : 0< y <f(x)\}$ when $f$ is positive, and for a general function, $\int_E f = \int_E f^+ - \int_E f^-$. The authors afterward show an equivalence similar to this one (as the supremum of simple functions less or equal than $f$). $\endgroup$ – guidoar Nov 2 '18 at 0:09

This is false in general.

For instance, if $f$ is any integrable function which is positive everywhere (e.g., $x \mapsto e^{-x^2}$), then this is false. This is because any simple function which is greater than $f$ will have infinite integral.

The requirement for your statement to be true is that $f$ is bounded and nonzero only on a set of finite measure.

This is because we then have that $f \leq M\cdot \mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}$. Therefore, $M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f \geq 0.$ Letting $g:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f$, we can take a sequence $s_n$ of simple functions which converge to $g$ monotonically. Defining $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n$, we have that $t_n \to M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g=f$.

By the monotone convergence theorem we have that $\int s_n \to M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int f$, and thus $$\int t_n=(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int s_n ) \to (M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))+\int f)=\int f.$$ By construction, $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n \geq M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g =f,$ and it is now easy to see that the result you state must hold.


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