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Assume $f:[a,b] \rightarrow \mathbb{R}$ is given. Prove or give a counter example.

If $f$ is integrable then $\vert f\vert $ is integrable.

My idea is let $f$ be integrable and $g(x) = \vert x\vert$. Then the composition $g(f) = \vert f\vert $ has same discontinuities as $f$, thus a zero set.

Eric
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    Welcome to MSE in the future please refer to the MathJax tutorial http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference –  Nov 11 '14 at 21:56
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    The two sets of discontinuities for $f$ and $|f|$ are not the same, For example, $f(0) = -1, f(x) = 1 $ when $x\in (0,1]$. However you can show if $f$ is continuous at $x$, then $|f|$ is also continuous at $x$, then apply your last argument. – Xiao Nov 12 '14 at 00:07
  • http://math.stackexchange.com/q/316090/ – Did Dec 29 '14 at 01:22

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Yes, your approach works. If $f$ is continuous at some point, then $|f|$ is also continuous there (shown here). So, you can use the theorem that says a function is Riemann integrable if and only if its set of discontinuity is a null set.

However, this theorem is a pretty heavy tool. A more natural approach is to observe that for every partition $P$ of the interval, the upper and lower sums (denoted $U$ and $L$ below) satisfy $$ U(|f|,P)-L(|f|,P)\le U(f,P)-L(f,P) \tag{1} $$ The proof of (1) is based on the inequality $||u|-|v||\le |u-v|$ applied to the values of $f$.

Community
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  • Can you clarify the proof of (1)? I'm trying to work through it at the moment based on another question, and keep getting $\epsilon$ inside the summations of the upper and lower sums, and it leads me to dead ends. – M T May 27 '15 at 15:08