# $\int_{-a}^{0}f(x) dx=\int_{0}^{a}f(x) dx$ for an even $f$

$f:\mathbb R \to \mathbb R$ integrable (but not necessarily continuous) on $\mathbb R$ function such that $f(x)=f(-x) , \forall x \in \mathbb R$.

I need to show that $\int_{-a}^{0}f(x) dx=\int_{0}^{a}f(x) dx, a \in \mathbb R$.

Intuitively the statement is quite clear, but I am missing an elegant way to show it formally.

Because $f$ is not continuous one can not resort to antiderivatives or use the substitution rule. One way is of course to prove the statement for a step function and then for any integrable $f$ but this seems to me disproportionally tedious for such intuitive statement. Am I overseeing something obvious here?

• – user223391 May 20 '16 at 19:04
• @ZacharySelk I fixed it so I don't use $u$-substitution. The proof is now a lot shorter. – Noble Mushtak May 20 '16 at 19:17

Assuming knowing measure theory (in particular the Lebesgue integral theory), one has the following short proof.

Suppose $c<0$. (Similarly one can deal with the case $c>0$.) Consider the linear transformation on $\mathbb{R}$. $T:x\mapsto -x$. Then $|\det T|=1$. Suppose $F\in L^1(\mathbb{R})$. Note in real analysis that we have $$\int_\mathbb{R}F(x)\ dx=|\det T|\int_\mathbb{R}F\circ T(x)\ dx,$$ namely in this particular case $$\int_\mathbb{R}F(x)\ dx= \int_\mathbb{R}F(-x)\ dx\quad\tag{*}$$ Now, assuming $f$ is an even function, let $$F(x)=f(x)1_{[0,-c]}(x)\quad x\in\mathbb{R}.$$ Then $$F(-x)=f(x)1_{[0,-c]}(-x)=f(x)1_{[c,0]}(x).$$

Now, apply $(*)$, noting that when $-\infty<a<b<\infty$, the Riemann integral $$\int_b^af(x)\ dx:=-\int_a^bf(x)\ dx=-\int_{[a,b]}f(x)\ dx=-\int_{\mathbb{R}}f(x)1_{[a,b]}(x)\ dx.$$ where $\int_{[a,b]}$ denotes the Lebesgue integral.

Let's use the definition of an integral to prove this (I'm using a very specific definition of the integral to make this easier): $$\int_a^b f(x)dx=\lim_{n \to \infty}\sum_{i=0}^n f\left(a+i\frac{b-a}{n}\right)\frac{b-a}{n}$$ Thus, we have: $$\int_{-a}^0 f(x)dx=\lim_{n \to \infty}\sum_{i=0}^n f\left(i\frac{-a}{n}\right)\frac{a}{n}$$ $$\int_0^a f(x)dx=\lim_{n \to \infty}\sum_{i=0}^n f\left(i\frac{a}{n}\right)\frac{a}{n}$$

Since $f$ is even $f\left(i\frac {-a} n\right)=f\left(i\frac a n\right)$, so clearly, these two limits are equal, making the integrals equal.

• Well written, +1 – user223391 May 20 '16 at 19:18

Let $\{t_0...t_n\}$ be a partition of $[0, a]$. To each such partition, we can, in a bijective manner, assign the partition $\{j_0...j_n\}$ to $[-a, 0]$ such that $j_i =-t_{n-i}$. It is easy to see, since $f$ is even, that with such partitions, the upper and lower Darboux sums will be the same for any partition. As this is a bijective correspondence, the upper and lower Darboux integrals will thus be the same as well.

Without loss of generality, assume that $f \geq 0$ and recall the definition of the Lebesgue integral:

$$\int_{A} f\, d\mu = \sup \left\{ \int_{A} g\, d\mu : 0\leq g \leq f \right\}$$

where the supremum is taken over simple functions $g$ which are smaller than or equal to $f$ in the set $A$ and vanish outside $A$. Let us denote

$$S_A = \left\{ \int_{A} g\, d\mu : 0 \leq g \leq f \right\}$$

where the simple functions $g$ are chosen in the sense explained above.

We are done if we show that

$$\int_{[0, a]} f\, d\mu = \int_{[-a, 0]} f\, d\mu$$

or $\sup S_{[0, a]} = \sup S_{[-a, 0]}$. To do this, we show that $S_{[0, a]} = S_{[-a, 0]}$. Indeed, let $x \in S_{[0, a]}$. By definition, there is a simple function $g \leq f$ vanishing outside $[0, a]$ such that

$$\int_{[0, a]} g \, d\mu = x$$

Let $g$ have canonical representation

$$g = \sum_{k=1}^n X_{E_k} a_k$$

and consider the function $h$ given by

$$h = \sum_{k=1}^n X_{-E_k} a_k$$

where $-A = \{ -x : x\in A \}$. Being the finite linear combination of the characteristic functions of measurable sets, $h$ is a simple function. Moreover, it vanishes outside $[-a, 0]$, and by evenness of $f$ we have that $h \leq f$ in the interval. It follows that we have

$$\int_{[-a, 0]} h\, d\mu \in S_{[-a, 0]}$$

On the other hand, by the definition of the Lebesgue integral, we have

$$\int_{[-a, 0]} h\, d\mu = \sum_{k=1}^{n} \mu(-E_k) a_k = \sum_{k=1}^{n} \mu(E_k) a_k = \int_{[0, a]} g\, d\mu = x$$

Therefore, $x \in S_{[-a, 0]}$, which shows that $S_{[0, a]} \subseteq S_{[-a, 0]}$. The argument can be repeated in the reverse direction to obtain the reverse inclusion, so that we have $S_{[0, a]} = S_{[-a, 0]}$, finishing the proof.

To extend the proof to functions which can take negative values as well, split the integral into its positive and negative parts, and apply this result on each individual part.