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Let $f$ be a continuous real-valued function on $(0,1]$. Show that \begin{equation} \int_{[0,1]} |f| \,dm = \lim_{n \rightarrow \infty} \int_{1/n}^1 |f(x)| \,dx \end{equation}

where the integral $\displaystyle{\int_{1/n}^1 |f(x)| \,dx}$ is the Riemann integral.

Here $m$ is the measure on $(0,1]$ satisfying $m(a,b) = b-a$.

My understanding of the formalities here is not very good. I am tempted to say that as $n \rightarrow \infty$ it follows that $1/n \rightarrow 0$ and so we have in the limit that:

\begin{equation} \lim_{n \rightarrow \infty} \int_{1/n}^1|f(x)|\,dx = \int_0^{\infty} |f(x)|\,dx = \int_{[0,1]}|f|\,dm \end{equation}

where the last equality follows as the Riemann integral exists, it must be equal to the Lebesgue integral.

I feel uneasy about this proof however as I have not used the hypothesis that $f$ is continuous on $(0,1]$. Can someone help with the formalities behind this proof?

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2 Answers

up vote 1 down vote accepted

W.l.o.g., assume that $f$ is nonegative to start with. The continuity is only needed to show that the function is integrable. To make your argument more formal, define $f_n(x)$ to be $f(x)$ if $x\in(1/n,1]$ and $f_n(x)=0$ otherwise. Then the sequence $f_n$ converges pointwise almost everywhere (everywhere but at the point $0$) to $f$ and is pointwise increasing, So $$\int f=\int \lim f_n=\lim_n \int f_n=\lim_n \int_{1/n}^n f$$ by the monotone convergence theorem. Then you can argue, as you did, that all functions involved are actually Riemann integrable.

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I assume you have already learned that Lebesgue and Riemann agree if the integrand is continuous and we integrate over a compact interval.

Let $$f_n(x) = \begin{cases}|f(x)|&\text{if }x\ge\frac1n\\0&\text{otherwise}\end{cases}$$ Then $\int_{[0,1]} f_n dm =\int_{[\frac1n,1]} f_n dm =\int_{\frac1n}^1 |f(x)|dx$ because we have a compact interval and a continuous function. The sequence $f_n$ is pointwise nondecreasing and hence $$\int_{[0,1]}|f|dm=\int_{[0,1]} \sup_{n\in\mathbb N} f_n dm=\sup_{n\in\mathbb N}\int_{[0,1]} f_n dm.$$ On the other hand we also have $$ \lim_{n\to\infty}\int_{\frac1n}^1 |f(x)|dx=\sup_{n\in\mathbb N}\int_{\frac1n}^1 |f(x)|dx$$ (also, because the sequence of the $\int_{\frac1n}^1 |f(x)|dx$ is nondecreasing), whence the claim.

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