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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.

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3 Answers 3

up vote 1 down vote accepted

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.

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Nice answer. It's worth pointing out that this is also true for the Riemann integral, and not just the Lebesgue one. It's not entirely trivial to know that if $f$ is integrable, then so is $f \chi_{\{f > \alpha\}}$. –  user61527 Jul 3 at 5:17
    
@T.Bongers Good point, I was somewhat sloppy about differentiating between integration of $f|[a,b]$ on $[a,b]$ and integration of $f\chi_{[a,b]}$ on $\mathbb R$. But note that $f\chi_{[a,b]}$ is measurable and $f$ is continuous on the compact and finite-measure set $[a,b]$ and hence $|f|$ attains its maximum here. It follows that $|f|\chi_{[a,b]}$ is integrable and thus so is $f\chi_{[a,b]}$. –  triple_sec Jul 3 at 5:49
    
Not very helpful. If you don't know the answer to the OP's question you probably don't know about Lebesgue integration on abstract measure spaces. And if you do, the above is probably quite obvious to you. –  jwg Jul 3 at 11:42
    
@jwg I agree that this proof is not the simplest one there is—nor did I intend it to be: I meant to emphasize how the result the OP requested a proof for carries over to a general setting. Indeed, many of the most beautiful and deep theorems/axioms in real analysis are mere generalizations of results that can be much more simply proven in certain less general settings: e.g. Urysohn's lemma, Tychonoff's theorem, Krein–Milman, the possibility of well-ordering, etc. I think this, too, is a nice example of how a very general result can be applied to a very specific problem. –  triple_sec Jul 4 at 4:59

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.

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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 $\alpha,\beta$.

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