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The total variation of differentiable function $f$ on the closed interval $[a,b]\subset\mathbb{R}$ is given by $$V_a^b(f)=\int_a^b|f'(x)|dx.$$ Does the same formula hold for the total variation of differentiable function on $\mathbb{R}$, i.e., can we say that the total variation of a function on $\mathbb{R}$ is equal to $\int_\mathbb{R}|f'(x)|dx$ (given that the last integral is finite)?

If the formula indeed holds, is there any book or other reference where it is proved?

Almost all the sources that I have checked deal with the variation on finite intervals paying almost no attention to variations on $\mathbb{R}$.

Perman
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1 Answers1

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I hope this clarifies a few things for the OP:

If $f$ is absolutely continuous on every compact interval of $\mathbb{R}$, then $f$ is of locally finite variation, $f'$ exits and is locally integrable and $$\begin{align} f(b)-f(a)&=\int^b_a f'(t)\,dt\\ V(f;a,b)&=\int_{(a,b]}|f'(t)|\,dt \end{align}$$

If in addition $f'\in L_1(\mathbb{R}$, then $f(-\infty):=\lim_{a\rightarrow-\infty}f(a)$ and $f(\infty)=\lim_{b\rightarrow\infty}f(b)$ exist, are finite, and $$\begin{align} f(x)&=f(-\infty)-\int_{(-\infty,x]}f'(t)\,dt\\ f(x)&=f(\infty)-\int_{(x,\infty)}f'(t)\,dt \end{align}$$ Furthermore, the measure $\mu_{f'}(A)=\int_Af'(t)\,dt$ is a Radon measure, its variation measure is given by $|\mu_{f'}|(A)=\int_A|f'(t)|\,dt$, and it has total finite variation $\|\mu_{f'}\|_{TV}=\int_\mathbb{R}|f'|$.

Mittens
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