# Proving that $\int f=\inf \left\lbrace\int g : g\text{ is a simple function such that } f\le g \right\rbrace$

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.

• Is this not immediate by definition? – bitesizebo Nov 1 '18 at 23:23
• @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 $f$). – guidoar Nov 2 '18 at 0:09

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.