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Show that if $f$ is integrable on $[a, b]$ then $\lvert f \rvert$ is also integrable. Hint: Show that $$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$
I have:
$$U (P , \lvert f \rvert) \ge U (P , f )$$ and $$ L(P , \lvert f \rvert) \ge L(P,f)$$
Thus,
$$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$

Martin Sleziak
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user72195
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2 Answers2

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Consider a partition $$P=\{x_{0},x_{1},x_{2},\dots,x_{n}\} $$ of $[a, b] $ and for each $i=1,2,\dots,n$ let us use the notation $$M_{i}(f) =\sup\, \{f(x) \mid x\in[x_{i-1},x_{i}]\},\, m_{i}(f)=\inf\,\{f(x) \mid x\in[x_{i-1},x_{i}]\} $$ Then we have by definition $$U(P, f) =\sum_{i=1}^{n}M_{i}(f)(x_{i}-x_{i-1}),\,L(P,f)=\sum_{i=1}^{n}m_{i}(f)(x_{i}-x_{i-1})$$ and hence $$U(P, f) - L(P, f) =\sum_{i=1}^{n}(M_{i}(f)-m_{i}(f))(x_{i}-x_{i-1})$$ It is easy to prove that $$M_{i} (f) - m_{i} (f) =\sup\, \{|f(x) - f(y) | : x, y\in[x_{i-1},x_{i}]\} $$ Now by triangle inequality we know that $$||f(x) |-|f(y) ||\leq |f(x) - f(y) |$$ Therefore it follows that $$M_{i} (|f|) - m_{i} (|f|) \leq M_{i} (f) - m_{i}(f) $$ which leads to the desired inequality $$U(P, |f|) - L(P, |f|) \leq U(P, f) - L(P, f) $$ From this inequality and criterion for Riemann integrability it is almost obvious that if $f$ is Riemann integrable on $[a, b] $ then $|f|$ is also Riemann integrable on $[a, b] $.

Paramanand Singh
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  • Hello, how can I prove that $M_i(f)-m_i(f)=\sup...$? Thanks! –  Apr 22 '21 at 17:10
  • @TheCalc: well, first note that both $f(x), f(y) $ lie in the interval $[m_i(f) , M_i(f) ]$ and hence the difference between them does not exceed $M_i(f) - m_i(f) $. And thus $M_i(f) - m_i(f) $ is an upper bound for all numbers of the form $|f(x) - f(y) |$. To show that it is indeed the least upper bound we can take $\epsilon$ with $0<\epsilon <M_i(f) - m_i(f) $ and choose $x, y$ such that $M_i(f)-\epsilon /2<f(x)\leq M_i(f) $ and $m_i(f) \leq f(y) <m_i(f) +\epsilon /2$. Then $M_i(f) - m_i(f) - \epsilon<|f(x) - f(y) |<M_i(f) - m_i(f) $. – Paramanand Singh Apr 22 '21 at 17:24
  • Thank you for your response! Why is $M-m$ is an upper bound for all numbers of the form $|f(x)-f(y)|$ and not just $f(x)-f(y)$? –  Apr 22 '21 at 17:43
  • And also, $M_i(f)-m_i(f)$ can be equal to zero, so $\epsilon \ge 0$ which is not good... If for instance, $f$ is constant. –  Apr 22 '21 at 17:47
  • @TheCalc: if $c, d$ lie in interval $[a, b] $ can you prove that $|c-d|\leq b-a$? If not then draw these points on number line and try to prove again. – Paramanand Singh Apr 22 '21 at 18:08
  • @TheCalc: if $M=m$ then $f$ is constant on that interval and the desired result holds. – Paramanand Singh Apr 22 '21 at 18:11
  • Thank you, what desired result holds? The whole statement about $M_{|f|}-m_{|f|} \le M_f-m_f$? –  Apr 23 '21 at 05:17
  • @TheCalc: if $f$ is constant on that interval $[x_{i-1},x_i]$ then the result $M_i(f) - m_i(f) =\sup, |f(x) - f(y) |$ holds because each side is $0$. – Paramanand Singh Apr 23 '21 at 05:34
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This answer shows that if $f$ and $g$ are Riemann integrable so is $h=\max\{f,g\}$. Use this with $g=-f$, then $h=|f|$.

Nota: In your post you "deduce" from $U (P , \lvert f \rvert) \geqslant U (P , f )$ and $ L(P , \lvert f \rvert) \geqslant L(P,f)$ that $U (P , \lvert f \rvert) − L(P , \lvert f \rvert) \leqslant U (P , f ) − L(P , f )$. This does not hold, in fact one can deduce no upper bound of $x-y$ from the hypothesis that $x\leqslant x'$ and $y\leqslant y'$. Instead one would need bounds such as $x\leqslant x'$ and $y\geqslant y'$, which indeed imply $x-y\leqslant x'-y'$.

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