Final answer:
The proof is based on the convergence of upper and lower sums of |f| when f is integrable on [a,b]. The absolute value of the integral of f from a to b is less than or equal to the integral of the absolute value of f due to the treatment of positive and negative areas.
Step-by-step explanation:
To prove that |f| is integrable if f is integrable on [a,b], we use the definition that a function is integrable if its upper and lower sums get arbitrarily close as the partitions of the interval [a,b] get finer. Since |f(x)| is always positive or zero, its upper and lower sums will also converge as those of f do, proving |f| is integrable. Moreover, the integral of f over [a,b] gives the net area between the function and the x-axis, which implies that | ∫ab f(x)dx | is less than or equal to ∫ab |f(x)|dx as it disregards the sign. This is evident by considering that the total positive area plus the absolute value of the total negative area is always greater than or equal to the net area.
For example, let's consider a function that produces an odd function xe-x² (odd times even is odd). The integral of an odd function over a symmetrical interval around the origin is zero since the positive area above the x-axis cancels out the negative area below it. However, the integral of |xe-x²| over the same interval would not be zero as it only considers the absolute values, resulting in a positive area.