Computing $\lim_{n\to\infty}\int_0^{\frac{n}{2}}\left(1-\frac{2x}{n}\right)^n d\lambda(x)$

Question

Let $\lambda$ be the Lebesgue measure in $\mathbb{R}$, then compute: $$\lim_{n\to\infty}\int_0^{\frac{n}{2}}\left(1-\frac{2x}{n}\right)^n d\lambda(x)$$

I thought of using the Monotone Convergence Theorem to pass the limit inside:

$$\lim_{n\to\infty}\int_0^{\frac{n}{2}}\left(1-\frac{2x}{n}\right)^n d\lambda(x)=\int_0^{\frac{n}{2}}\lim_{n\to\infty}\left(1-\frac{2x}{n}\right)^n d\lambda(x)=\int_0^{\frac{n}{2}}e^{-2x}\,d\lambda(x)$$

Since the function $e^{-2x}$ is continuous then: $$\int_0^{\frac{n}{2}}e^{-2x}\,d\lambda(x)=\int_0^{\frac{n}{2}}e^{-2x}dx=-\frac{e^{2x}}{2}|_0^{\frac{n}{2}}=?$$

Questions:

Now the $\frac{n}{2}$ gets unaffected by the limit if I apply the Monotone Convergence Theorem. What am I doing wrong? How should I compute the integral?

Answer 1

The function you should apply MCT to is $f_n(x) = 1_{(0,n/2)}(1-2x/n)^n$ where $1$ is the indicator function. Note that in the MCT the underlying set of integration is the same throughout. Here you may apply MCT as you have done, realizing that $1_{(0,n/2)}\uparrow 1$ and $(1-2x/n)^n\uparrow e^{-2x}$.

Answer 2

Are we using a rocket launcher to take down a cockroach? Define $$u=1-\frac{2x}{n}$$ Then the integral becomes $$\int_0^1 x^n \frac{n}{2}dx=\frac{n}{2(n+1)}\rightarrow .5$$