For which $f$ do we have $\int_{−∞}^{+∞} f (x + m) \, \mathrm{d}x = \int_{−∞}^{+∞} f (x) \, \mathrm{d}x$?

Question

So, I want to believe that there is a theorem like so:

$$∀ m. \int_{−∞}^{+∞} f (x + m) \, \mathrm{d}x = \int_{−∞}^{+∞} f (x) \, \mathrm{d}x$$

proof by infinity

Assume $∀ x. ∞ + x = ∞$.

$$\int_{−∞}^{+∞} f (x + m) \, \mathrm{d}x \overset{\text{substitution}}= \int_{−∞ + m}^{+∞ + m} f (x) \, \mathrm{d}x \overset{\text{assumption}}= \int_{−∞}^{+∞} f (x) \, \mathrm{d}x$$

∎ Slick! But $∞$ is not a real number. I wonder if this can be formalized with some fancy «extended real numbers» kind of theory.

proof by principal value

Assume $∀ m. \lim_{a \to ∞}\int_{a − m}^{a + m} f (x) \, \mathrm{d}x = 0$.

$$\begin{align} &\int_{−∞}^{+∞} f (x + m) \, \mathrm{d}x\\ \small{\text{(principal value)}}= \lim_{a \to ∞}&\int_{−a}^{+a} f (x + m) \, \mathrm{d}x\\ \small{\text{(substitution)}}= \lim_{a \to ∞}&\int_{−a + m}^{+a + m} f (x) \, \mathrm{d}x\\ \small{\text{(split at $a − m$)}}= \lim_{a \to ∞}&\int_{−(a − m)}^{+(a - m)} f (x) \, \mathrm{d}x + \int_{a − m}^{a + m} f (x) \, \mathrm{d}x\\ \small{\text{(linearity of limit)}}= \lim_{a \to ∞}&\int_{−(a − m)}^{+(a - m)} f (x) \, \mathrm{d}x + \lim_{a \to ∞}\int_{a − m}^{a + m} f (x) \, \mathrm{d}x\\ \small{\text{(assumption)}}= \lim_{a \to ∞}&\int_{−(a − m)}^{+(a - m)} f (x) \, \mathrm{d}x\\ \small{\text{(principal value)}}= &\int_{−∞}^{+∞} f (x) \, \mathrm{d}x\\ \end{align}$$

∎ Nice! And it is plausible that a bounded function with finite total weight will have infinitely light tails. But how do I formalize this?

my question

So, for which $f$ is the above statement true? Is there a standard reference?

I found something called Glasser's master theorem, but it is formulated differently in different sources so I am not sure if it applies. I have no idea how to prove it anyway.

Answer 1

Improper Riemann integral: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that the improper Riemann integral $\int_{-\infty}^\infty f(x) dx$ exists. That means for every $a\in \mathbb{R}$ we have that the limits $$ \lim_{R\rightarrow \infty} \int_a^R f(x) dx, \quad \lim_{R\rightarrow -\infty} \int_R^a f(x) dx $$ exist. In this case the improper Riemann integral is equal to $$ \int_{-\infty}^\infty f(x) dx = \lim_{R\rightarrow \infty} \int_{-R+a}^{R+a} f(x) dx $$ for any fixed $a\in \mathbb{R}$.

This reduces the problem to show the change of variables for a proper Riemann integral. We have $$ \int_{-R}^R f(x+m) dx = \lim_{N\rightarrow \infty} \sum_{j=1}^N \frac{2R}{N} f(m-R+ j \frac{2R}{N}) = \int_{-R+m}^{R+m} f(x) dx. $$ Putting things together, we get $$\int_{-\infty}^\infty f(x+m) dx = \lim_{R\rightarrow \infty} \int_{-R}^R f(x+m) dx = \lim_{R\rightarrow \infty} \int_{-R+m}^{R+m} f(x) dx = \int_{-\infty}^\infty f(x)dx.$$

Lebesgue integral: Let's assume now that $f\in L^1(\mathbb{R})$ (i.e. $f$ is Lebesgue integrable). The standard way to prove identities is to proceed as follows:

1.) Show that it holds true for indicator functions $f(x)=1_A(x)$.

2.) By linearity it holds true for $f$ being a simple function.

3.) Using monotone convergence it holds true for all nonnegative integrable function.

4.) Splitting into negative and positive part and using again linearity it holds for all real, integrable functions.

5.) Splitting into real and imaginary part it holds for all integrable functions.

Thus, the only thing one really needs to check is that our identity holds true for indicator functions. This means, the statement we want to prove is that $$ \mathcal{L}(A) = \int_{\mathbb{R}} 1_A(x) dx = \int_{\mathbb{R}} 1_A(x+m) dx = \mathcal{L}(A-m), $$ where $\mathcal{L}$ denotes the Lebesgue measure on $\mathbb{R}$ and $$ A-m = \{a-m \in \mathbb{R} \ : \ a\in A\}.$$ This last equality is the fact that the Lebesgue measure is translation-invariant (see here Is Lebesgue measure translation invariant? for a proof).

Cauchy's principal value: Finally, we can interpret your integrals as a principal value with singularities at infinity. Namely, if you think of this expression as $$ \int_{-\infty}^\infty f(x) dx = \lim_{R\rightarrow \infty} \int_{-R}^R f(x)dx. $$ In this case your identity does not hold. Consider the function $f(x)=x$. Then we have $$ \lim_{R\rightarrow \infty} \int_{-R}^R x dx = \lim_{R\rightarrow \infty} \int_{-R}^R x dx = \lim_{R\rightarrow \infty} 0 =0. $$ On the other hand, we have for $m>0$ $$ \lim_{R\rightarrow \infty} \int_{-R}^R f(x+m) dx = \lim_{R\rightarrow \infty} \int_{-R}^R (x+m) dx = \lim_{R\rightarrow \infty} \int_{-R}^R m \, dx = \infty. $$