# How do I show that if $f$ is bounded and integrable on $\mathbb{R}$, then $g(t) = \int_t^{t+1} f(x) dx$ is continuous?

Usually functions of the form of $g(t)$ tell me I should use the fundamental theorem of calculus, but I don't think that applies here because I'm not given that $f$ is continuous.

I know from the definition of continuity that I need to show for some $\epsilon$ and some $x \in \mathbb{R}$, there exists a $\delta$ such that if $|x - y| < \delta$, then $|g(x) - g(y)| < \epsilon$.

$f$ is bounded, so there is some $M$ so that $f(t) < M$ for all real numbers $t$. Also, since $f$ is integrable, I can find two sequences of partitions $P_{x,k}$ and $P_{y,k}$ so that

1. $\lim_{k \to \infty} |P_{x,k}| = 0$ and $\lim_{k \to \infty} |P_{y,k}| = 0$, and

2. $\int_x^{x+1} f = \lim_{k \to \infty} U(f, P_{x,k})$ and $\int_y^{y+1} f = \lim_{k \to \infty} U(f, P_{y,k})$.

For any $k \in \mathbb{N}$, the sequences $P_{x,k}$ and $P_{y,k}$ may have different numbers of points, so I used $N_x$ and $N_y$ for these numbers of points, for some arbitrary $k$.

Ploughing onwards, I tried to work out what my $\delta$ should be be working backwards from the definition of continuity, but that gave me this:

\begin{align} |g(x) - g(y)| &= \left| \int_x^{x+1} f(u)du - \int_y^{y+1} f(u)du \right| \\ &= \left| \lim_{k \to \infty} (U(f, P_{x,k} - L(f, P_{x,k}) - \lim_{k \to \infty} (U(f, P_{y,k} - L(f, P_{y,k}) \right| \\ &= \left| \lim_{k \to \infty} \sum_{i=0}^{N_x - 1} (M_{x,i} - m_{x,i})(x_{i+1} - x_i) - \lim_{k \to \infty} \sum_{i=0}^{N_y - 1} (M_{y,i} - m_{y,i})(y_{i+1} - y_o) \right| \\ &< \left| 2M \lim_{k \to \infty} \sum_{i=0}^{N_x - 1} (x_{i+1} - x_i) - 2M \lim_{k \to \infty} \sum_{i=0}^{N_y - 1} (y_{i+1} - y_i) \right| \\ &= \left| 2M \lim_{k \to \infty} (x_{N_x} - x_0) - 2M \lim_{k \to \infty} (y_{N_y - 1} - y_0) \right| \\ &= \left| 2M \lim_{k \to \infty} (1) - 2M \lim_{k \to \infty} (1) \right| \\ &= 0 \end{align} Sadly this didn't give me a $\delta$, and I'm not sure where I used the assumption that $|x - y| < \delta$, whatever that $\delta$ might be.

Did I take a wrong turn somewhere?

• You don't need the Riemann sum to do it, only need to estimate variation of integral around $t, t+\Delta t$ and.$t+1, t+1+\Delta t$. See the answer below. – Math Wizard Jun 1 '15 at 0:57

Let $M\ge 0$ such that $$|f(x)|\le M \quad \forall x\in \mathbb{R}.$$ For every $t,s \in \mathbb{R}$ we have: \begin{eqnarray} |g(s)-g(t)|&=&\left|\int_t^{t+1}f(x)\,dx-\int_s^{s+1}f(x)\,dx\right|\\ &=&\left|\int_t^0f(x)\,dx+\int_0^{t+1}f(x)\,dx-\int_s^0f(x)\,dx-\int_0^{s+1}f(x)\,dx\right|\\ &=&\left|\int_t^sf(x)\,dx+\int_{s+1}^{t+1}f(x)\,dx\right|\\ &\le&\int_{\min\{t,s\}}^{\max\{t,s\}}|f(x)|\,dx+\int_{1+\min\{t,s\}}^{1+\max\{t,s\}}|f(x)|\,dx\\ &\le&\int_{\min\{t,s\}}^{\max\{t,s\}}M\,dx+\int_{1+\min\{t,s\}}^{1+\max\{t,s\}}M\,dx\\ &=&M(\max\{t,s\}-\min\{t,s\})+M(1+\max\{t,s\}-1-\min\{t,s\})\\ &=&M|t-s|+M|t-s|=2M|t-s| \end{eqnarray} Hence $g$ is Lipschitz-continuous on $\mathbb{R}$.
I think you're making it more difficult than it has to be. We have $$|g(x)-g(y)|=\left|\int_x^{x+1}f(t)dt -\int_y^{y+1}f(t)dt\right|\\=\left|\int_x^yf(t)dt-\int_{x+1}^{y+1}f(t)dt\right|$$ Now use that $f$ is bounded.
Suppose $\Delta t<1, |f(x)|<M$ \begin{align} |g(t+\Delta t)-g(t)|&=\left|\int_{t+\Delta t}^{t+\Delta t+1}f(x)dx-\int_{t}^{t+1}f(x)dx\right| \\ &=\left|\int_{t+\Delta t}^{t+1}f(x)dx+\int_{t+1}^{t+\Delta t+1}f(x)dx-\int_{t}^{t+\Delta t}f(x)dx-\int_{t+\Delta t}^{t+1}f(x)dx\right| \\ &=\left|\int_{t+1}^{t+\Delta t+1}f(x)dx-\int_{t}^{t+\Delta t}f(x)dx\right| \\ &\leqslant |M|\left(\left|\int_{t+1}^{t+\Delta t+1}dx\right|+\left|\int_{t}^{t+\Delta t}dx\right|\right) \\ &=2M|\Delta t| \\ &\to0 \hspace{3 mm} \text{as } t\to0 \end{align}
So $g(t)$ is continuous.