# real analysis – Solution Check: Determine if \$lim_{nrightarrow infty} int_X e^{-nf(x)} dmu(x) \$ exists.

I would appreciate if I could get this proposed solution checked for correctness and rigor. Here is the question:

Let $$(X,mathcal{M},mu)$$ be a finite measure space. Let $$f:Xrightarrow (0,infty)$$ be measurable. If the following limit exists, determine its value:

$$lim_{nrightarrow infty} int_X e^{-nf(x)} dmu(x).$$

If it does not, explain. In either case, justify your claims.

Here is my solution:

Since f is a non negative function notice that
$$e^{-nf(x)} leq e^{-n(0)} =e^0 =1.$$

Since $$X$$ is a finite measure space we have
$$int_X 1 dmu(x)= mu(X)

for some $$Min (0,infty)$$. Therefore by the Dominated Convergence theorem we have
$$lim_{nrightarrow infty} int_X e^{-nf(x)} = int_X lim_{nrightarrow infty} e^{-nf(x)}.$$

Suppose $$f = 0$$ a.e., then
$$int_X lim_{nrightarrow infty} e^{-nf(x)} = int_X 1 dmu(x) =mu(X).$$

On the other hand, if $$f neq 0$$ a.e., then
$$int_X lim_{nrightarrow infty} e^{-nf(x)} = int_X 0 dmu(x) =0.$$

Therefore the limit exists and we see that

$$lim_{nrightarrow infty} int_X e^{-nf(x)} dmu(x)= begin{cases} mu(X) & text{if } f= 0 text{ a.e.}\ 0 & text{if } fneq 0 text{ a.e.} end{cases}$$