# Integral of a function by Lebesgue measure and monotony convergence

Let be $$f:[0, 1] \rightarrow \mathbb{R}$$ a function such that $$\int_{[0,1]}|f|<+\infty$$. Calculate: $$\lim_{n \rightarrow +\infty}\frac1n\int_0^1log(1+e^{nf(x)})dx$$ I wanted to either use the dominated convergence theorem, but I dont know how to find a function that is upperbound Lebesgue integrable function, or the monotony convergence theorem, but I dont know how to prove the monotony. Any help?

• I think that it really depend on $f$. For example, if $f\equiv 0$, then the limit is 0, however, if $f\equiv 1$, the limit will be $\infty$.
– Surb
Mar 11, 2022 at 14:29
• How can I calculate the integral as function of $f$? Mar 11, 2022 at 15:00

First, consider the following:

Where $$f(x)>0$$, the expression $$\ln(1+e^{nf(x)})$$ is almost $$\;nf(x)$$.

Where $$f(x)=0$$, you get $$\ln(1+e^{nf(x)})=\ln(2)$$.

Where $$f(x)\leq0$$ instead, $$\ln(1+e^{nf(x)})$$ gets close to $$0$$.

This helps us to estimate $$\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x$$.

$$\\$$

Let $$S=\{t\in[0,1]:f(t)>0\}$$.

If $$S$$ has positive measure, then automatically $$\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x=+\infty$$.

So we may as well now assume $$S$$ is a null set.

$$\\$$

Let $$T=\{t\in[0,1]:f(t)=0\}$$.

You can see that $$\displaystyle\int_T\ln(1+e^{nf(x)})\;\text{d}x=\ln(2)\mu(T)$$ for all $$n\in\mathbb{Z}^+$$.

$$\\$$

Let $$U=\{t\in[0,1]:f(t)<0\}$$

As for $$\displaystyle\int_U\ln(1+e^{nf(x)})\;\text{d}x$$, how do we know that it tends to $$0$$ as $$n$$ tends to infinity?

This is an application of a standard trick: look at $$U_r=\{t\in[0,1]:f(t)<-r\}$$ where $$r>0$$.

Obviously for a fixed $$r$$,

$$\displaystyle\int_{U_r}\ln(1+e^{nf(x)})\;\text{d}x \leq\mu(U_r)\ln(1+e^{-rn})\xrightarrow[n \to \infty]{} 0$$.

So, given that $$\displaystyle\int_U\ln(1+e^{nf(x)})\;\text{d}x=\int_{U_r}\ln(1+e^{nf(x)})\;\text{d}x+\int_{U\setminus U_r}\ln(1+e^{nf(x)})\;\text{d}x$$,

by varying $$r$$, we can make the second summand arbitrarily small, because $$\ln(1+e^{nf(x)})<1$$ on $$x\in U$$, and $$\displaystyle\lim_{r\to0^+}\mu(U\setminus U_r)=0$$;

varying $$n$$ after we vary $$r$$, we can make the first summand tend to $$0$$.

So $$\displaystyle\lim_{n\to\infty}\int_U\ln(1+e^{nf(x)})\;\text{d}x=0$$.

$$\\$$

In conclusion,

$$\displaystyle\lim_{n \rightarrow +\infty}\int_0^1\ln(1+e^{nf(x)})\;\text{d}x=\begin{cases} +\infty & \text{if } \mu(S)>0 \\ \ln(2)\mu(T) & \text{if } \mu(S)=0 \end{cases}$$

• So, there is no need to use the monotone convergence theorem or the dominated convergence theorem? Also can you please explain with more details the first part? Why the fact that $log(1+e^{nf(x)})$ is almost $nf(x)$ implies that the integral of $log(1+e^{nf(x)})$ diverges to $+\infty$? Mar 11, 2022 at 15:26
• If $S$ has positive measure, then there is some $r>0$ such that $S_r=\{t\in[0,1]:f(t)>r\}$ has positive measure. For such a value of $r$, you get $\mu(S_r)\ln(1+e^{nr})\to\infty$ Mar 11, 2022 at 15:32