Let
$\,\mathrm{g}\left(x\right) =
\left\lbrace\begin{array}{rl}
\,\mathrm{f}\left(x\right)\,, & x \in \mathbb{Q}
\\[1mm]
-\,\mathrm{f}^{2}\left(x\right)\,, & x \not\in \mathbb{Q}
\end{array}\right.
\quad$
where $\,\mathrm{f}\left(x\right)$ continuous at $\left[0,1\right]$. Find a formula to
evaluate
$\displaystyle{\left(\mbox{L}\right)\int_{0}^{1}\,\mathrm{g}\left(x\right)
\,\mathrm{d}x}$ and find $\left(\mbox{L}\right)$ for
$\,\mathrm{f}\left(x\right) = \,\sqrt[3]{\, x\,}\,$.
$\left(\mbox{L}\right)$ denotes Lebesgue-integral.
I don't even know how to start, this is for exam I have on Tuesday.
I'm not familiar with Lebesgue-integral, can you help me please ?.
Answer
The Lebesgue integral doesn't care about sets of measure zero. So, you can change $g$ to whatever you want on $[0,1] \cap \mathbb{Q}$ (in particular, the natural choice is $-f^2(x)$) and still get the same value.
So, your integral is equal to $\int_0^1 -f^2$, which you can easily evaluate.
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