I'm trying to prove the next proposition:
Let $f:R\subset\mathbb{R}^n\rightarrow\mathbb{R}$ bounded such as $f(x)\geq c>0$ for all $x\in R,$ where $R$ is a closed rectangle. If $f^2$ is Riemann integrable over $R$ then $f$ also is Riemann integrable .
I'm lost. My attemp begings defining a new function $f(x)-c\geq0,$ so $\sup\{(f(x)-c)^2:x\in R_{i}\}=\sup^2\{f(x)-c:x\in R_{i}\},$ where $R_{i}$ are the rectangles induced by the partition, and a similar expression with the infimum. I use the integrability of $f^2$ to try bounding the difference between superior and inferior sums of $f$, but the inequalities are useless because of properties of supremum and infimum.
I'd aprecciate any kind of help. Thanks in advance.
Best Answer
Let $\varepsilon>0$. Since $f^2$ is Riemann integrable in $R$, there exists a partition $P$ of $R$ into rectangles $R_1,\dots R_n$ such that $$U_{f^2}(P)-L_{f^2}(P)<2c\varepsilon,$$ where $U_{f^2}(P)$ is the upper sum and $L_{f^2}(P)$ is the lower sum for $f^2$, for the partition $P$. Note now that, if $x,y\in R_i$, $$f(x)-f(y)=\frac{f^2(x)-f^2(y)}{f(x)+f(y)}\leq\frac{f^2(x)-f^2(y)}{2c}\leq\frac{1}{2c}\left(\sup_{R_i}f^2-\inf_{R_i}f^2\right),$$ which shows that $$\sup_{R_i}f-\inf_{R_i}f\leq \frac{1}{2c}\left(\sup_{R_i}f^2-\inf_{R_i}f^2\right).$$ We now add for $i=1,\dots n$, to obtain that $$U_f(P)-L_f(P)\leq\frac{1}{2c}\left(U_{f^2}(P)-L_{f^2}(P)\right)<\varepsilon,$$ therefore $f$ is Riemann integrable.