When does the integral preserve strict inequalities? Hi everyone: Suppose that $f(t)$ is a continuous function on $[a,b]$, where we also have 
$$\alpha<f<\beta.$$
 Under which condition(s) do the above strict inequalities are preserved by taking the integral: 
$$\alpha<\frac{1}{b-a}\int_{a}^{b}f(t)dt<\beta?$$ Thanks for your reply.
 A: Due to compactness and continuity, the function will have maximum and minimum, which are strictly better bound than $\alpha,\beta$. Hence there is already an inequality due to that max/min bound, thus a strict inequality for $\alpha,\beta$.
A: In fact, something way more general holds. Let $(X,\mathscr M,\mu)$ be an arbitrary measure space and $f:X\to\mathbb R$ a measurable function (that is $f^{-1}(U)\in\mathscr M$ for any Borel set $U\subseteq \mathbb R$). If


*

*$E\in\mathscr M$,

*$\mu(E)>0$, and

*$f(x)>0$ for all $x\in E$,


then $$\int_E f(x)\,\mathrm d\mu(x)>0.$$
Proof: Define $E_1\equiv\{x\in E\,|\,f(x)>1\}$ and for each $n\in\mathbb N$, $n\geq 2$, define $$E_n\equiv\left\{x\in E\,\Big|\,\frac{1}{n}<f(x)\leq\frac{1}{n-1}\right\}.$$
Since $f(x)>0$ for all $x\in E$, it follows that $E=\bigcup_{n=1}^{\infty} E_n$ and the members of the union are disjoint. Therefore, $$0<\mu(E)=\sum_{n=1}^{\infty} \mu(E_n),$$
so it follows that $\mu(E_n)>0$ for at least one $n\in\mathbb N$. Since $E_n\subseteq E$, it follows that $$\int_Ef(x)\,\mathrm d\mu(x)\geq\int_{E_n}f(x)\,\mathrm d\mu(x)\geq \int_{E_n}\frac{1}{n}\,\mathrm d\mu(x)=\frac{\mu(E_n)}{n}>0,$$
proving the claim.

In your case, $X=\mathbb R$, $\mathscr M$ is the Borel $\sigma$-algebra on $\mathbb R$, and $\mu$ is the Lebesgue measure. The function $x\mapsto f(x)-\alpha$ is continuous (and hence measurable) and strictly positive on $E=[a,b]$. Finally, $\mu([a,b])=b-a>0$. Hence,
\begin{align*}
0<&\int_{[a,b]}[f(x)-\alpha]\,\mathrm d x=\int_{[a,b]} f(x)\,\mathrm dx-\int_{[a,b]}\alpha\,\mathrm dx=\int_{[a,b]} f(x)\,\mathrm dx-\alpha\mu([a,b])\\
=&\int_{[a,b]} f(x)\,\mathrm dx-\alpha(b-a).\end{align*}
Your claim follows by simple rearrangement and the reasoning for the function $x\mapsto\beta-f(x)$ is totally analogous.
A: Always. We can write
\begin{align*}
\int_a^b f(t) dt &= \int_a^b \alpha dt + \int_a^b f(t) - \alpha dt \\
&= \alpha (b - a) + \int_a^b f(t) - \alpha dt \\
&> \alpha (b - a)
\end{align*}
because the integral of a strictly positive function over a (non-degenerate) interval is strictly positive. An identical argument works for the other inequality.
