# Prove that$|\int_{D}f(x)dx|\leq\int_{D}|f(x)|dx$.

Let $$D$$ be a subset in $$\Bbb{R}^n$$ and $$f:D\to\Bbb{C}$$. Prove that$$|\int_{D}f(x)dx|\leq\int_{D}|f(x)|dx$$.

If I can compare $$\pm f$$ with $$|f|$$, then the proof may be easy. But $$f$$ is a complex-valued function. Or I have to prove this by triangle inequality? I have no idea how to do this.

• – Martin R Sep 16 at 17:09

You can generalize that "compare $$\pm f$$ to $$|f|$$" thing to complex-valued functions: If $$z\in\Bbb C$$ then $$z=re^{it}$$ as usual; so there exists $$\alpha$$ with $$|\alpha|=1$$ and $$\alpha z\ge0$$.

Choose $$\alpha$$ with $$|\alpha|=1$$ and $$\alpha \int f\ge0$$; note that $$\left|\int f\right|=\alpha\int f$$. Let $$g=\Re(\alpha f).$$Then $$g\le|f|$$, so $$\left|\int f\right|=\int g\le\int|f|.$$

• sorry...what does it mean of kappa? – Maggie Sep 17 at 12:48
• @Maggie I have no idea what you're asking - there is no $\kappa$ in what I wrote... – David C. Ullrich Sep 17 at 13:26
• ohoh..I read it wrong. That is the real part of $\alpha f$. I've got it. thanks!! – Maggie Sep 18 at 13:20

A good way to approach this is to remember how (Riemann) integration is defined. Let's do the easy case, of a function from R to R.

A partition P of [a,b] is simply a way of cutting up the domain. E.g. A partition of $$[0,1]$$ could be made with the points $$0, 0.1, 0.2, ... 0.9, 1.0$$

Our lower sum just takes a partition and adds up the smallest value of f on each section of the partition, adjusted for the width of the section. The upper sum does the same, but adds the largest value of x on each section, adjusted for the width of the section.

In my clumsily drawn illustration above, the lower sum for the partition below is the sum of the area of the blue rectangles, while the upper partitions is the sum of the areas of the blue and black rectangles.

The key idea is that the integral is defined as the limit of the Lower sums and upper sums as the width of the intervals in the partition tends to 0. In fact, this is how we define the integral.

So, to prove your statement, we can write out the integral as a limit of upper sums, or of lower sums, of partitions whose width tends to $$0$$. We can then show that the inequality holds for every single term in the sequence of partitions - as these are finite sums, this literally is the triangle inequality. Finally, we use that if $$a_i \leq b_i$$ for all i, and the limit exists, then $$a \leq b$$ where a is the limit of $$a_i$$ and $$b$$ the limit of $$b_i$$

If you would like to understand the definition of the Riemann integral better, you can take a look at page 46 of these notes https://dec41.user.srcf.net/notes/IA_L/analysis_i.pdf