If $f$ is a Riemann integrable function on $[a,b]$, how can I show that the positive part of $f$, $f^+$, and the negative part of $f$, $f^-$, are also Riemann integrable?

Based off of this, how can I use these to prove that |$f$| is integrable and then $$ \left|\int_a^b f \right| \le \int_a^b \left|f \right|$$

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    $\begingroup$ If you know that $|f|$ is integrable, you can use $f^+=\cfrac{f+|f|}{2}$. $\endgroup$ – xavierm02 Nov 22 '13 at 21:37

I see that you want the integrability of $|f|$ as a conclusion and hence this fact can not be used in the proof of integrability of $f^{+}$ and $f^{-}$. However note that $f^{+} = (f + |f|)/2$. Let $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ denote a partition of $[a, b]$ and let us use the following notation: $$M_{k}(f) = \sup\{f(x)\mid x\in [x_{k - 1}, x_{k}]\}$$ and $$m_{k}(f) = \inf\{f(x)\mid x\in [x_{k - 1}, x_{k}]\}$$ Since $f$ is given to be Riemann-integrable on $[a, b]$, for any $\epsilon > 0$ there exists a partition $P$ of $[a, b]$ such that $$\sum_{k = 1}^{n}\{M_{k}(f) - m_{k}(f)\}(x_{k} - x_{k - 1}) < \epsilon$$ Now using properties of supremum and infimum one can show that $$M_{k}(f^{+}) = \dfrac{M_{k}(f) + M_{k}(|f|)}{2},\, m_{k}(f^{+}) = \dfrac{m_{k}(f) + m_{k}(|f|)}{2}$$ and therefore $$M_{k}(f^{+}) - m_{k}(f^{+}) = \dfrac{M_{k}(f) - m_{k}(f)}{2} + \dfrac{M_{k}(|f|) - m_{k}(|f|)}{2}$$ Also we have the general inequality $||f(x)| - |f(y)|| \leq |f(x) - f(y)|$ from which it follows that $M_{k}(|f|) - m_{k}(|f|) \leq M_{k}(f) - m_{k}(f)$. Thus we see that $$M_{k}(f^{+}) - m_{k}(f^{+}) \leq M_{k}(f) - m_{k}(f)$$ and therefore $$\sum_{k = 1}^{n}\{M_{k}(f^{+}) - m_{k}(f^{+})\}(x_{k} - x_{k - 1}) \leq \sum_{k = 1}^{n}\{M_{k}(f) - m_{k}(f)\}(x_{k} - x_{k - 1}) < \epsilon$$ and therefore the function $f^{+}$ is also Riemann-integrable on $[a, b]$. Same way we can do for $f^{-}$.

Also in the last paragraph, we have noted that $M_{k}(|f|) - m_{k}(|f|) \leq M_{k}(f) - m_{k}(f)$ which implies $$\sum_{k = 1}^{n}\{M_{k}(|f|) - m_{k}(|f|)\}(x_{k} - x_{k - 1}) \leq \sum_{k = 1}^{n}\{M_{k}(f) - m_{k}(f)\}(x_{k} - x_{k - 1}) < \epsilon$$ ans thus $|f|$ is Riemann-integrable on $[a, b]$.

The above derivation looks too much symbolic (as most of the proofs/derivation in real analysis) and for a beginner this may seem as just some fashionable symbol shunting. Hence it is important to have an informal version. The Riemann integrability depends upon making the difference between lower and upper sums as small as possible. The upper and lower sums themselves depend on the supremum and infimum of the function under consideration. So in effect we need to minimize the difference between infimum and supremum of a function for most part of the interval. Now in the above proof we have shown that the difference between sup and inf of $f^{+}$ does not exceed the difference between those of $f$. Since $f$ is Riemann-integrable the difference between sup and inf of $f$ is as small as desired for most part of the interval. It therefore follows that the difference between sup and inf of $f^{+}$ is as small as needed to establish its Riemann-integrability. The same technique has been used for $|f|$.

Establishing the integral inequality is not that difficult. There are two approaches both of which I will describe informally. First note that a Riemann integral is limit of Riemann sums as we make the partition of interval finer and finer (norm of partition tending to zero). Since we know that $|x + y| \leq |x| + |y|$, it follows that absolute value of a Riemann sum of $f$ is less than or equal to the corresponding Riemann sum of $|f|$. Hence their limits also follow the same inequality i.e. $$\left|\int_{a}^{b}f(x)\,dx\right| \leq \int_{a}^{b}|f(x)|\,dx$$

Another approach is to follow the obvious inequality $-|f(x)| \leq f(x) \leq |f(x)|$. Applying integral signs preserves inequalities (because integral of a non-negative function is non-negative) and hence we get $$-\int_{a}^{b}|f(x)|\,dx \leq \int_{a}^{b}f(x)\,dx \leq \int_{a}^{b}|f(x)|\,dx$$ and this means that $$\left|\int_{a}^{b}f(x)\,dx\right| \leq \int_{a}^{b}|f(x)|\,dx$$


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