# Absolute Value Integration

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• November 11th 2008, 10:23 AM
Coda202
Absolute Value Integration
Show that:

|f - g| is integrable.
• November 11th 2008, 10:37 AM
HallsofIvy
Quote:

Originally Posted by Coda202
Show that:

|f - g| is integrable.

Well, it isn't in general! If you forgot to include the hypotheses that f and g are themselves integrable, then you can use the fact that |f- g|= f- g as long as f(x)> g(x) and |f- g|= g- f as long as f(x)< g(x).
• November 11th 2008, 11:29 AM
ThePerfectHacker
Quote:

Originally Posted by Coda202
Show that:

|f - g| is integrable.

If $h_1,h_2$ are integrable then $h_1\pm h_2$ are integrable.
And if $h_3$ is integrable then $|h_3|$ is integrable.

Therefore, if $f,g$ are integrable then $f-g$ is integrable and that implies $|f-g|$ is integrable.